Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions

We are given the rule for the sequence

$$u_n = 2n - 1$$

where $$n$$ is a positive integer (that is, $$n = 1, 2, 3, \ldots$$). Let us substitute successive natural-number values of $$n$$ and calculate the corresponding terms.

1. For $$n = 1$$

   $$u_1 = 2 \times 1 - 1 = 2 - 1 = 1$$

2. For $$n = 2$$

   $$u_2 = 2 \times 2 - 1 = 4 - 1 = 3$$

3. For $$n = 3$$

   $$u_3 = 2 \times 3 - 1 = 6 - 1 = 5$$

4. For $$n = 4$$

   $$u_4 = 2 \times 4 - 1 = 8 - 1 = 7$$

5. For $$n = 5$$

   $$u_5 = 2 \times 5 - 1 = 10 - 1 = 9$$

6. For $$n = 6$$

   $$u_6 = 2 \times 6 - 1 = 12 - 1 = 11$$

The first six terms are therefore

$$1,\;3,\;5,\;7,\;9,\;11\ldots$$

Notice that each term is an odd number and that the difference between any two consecutive terms is

$$3 - 1 = 2, \; 5 - 3 = 2, \; 7 - 5 = 2, \; \text{and so on.}$$

Hence the sequence lists all positive odd integers in increasing order, and the n^th term is always given by $$u_n = 2n - 1$$, as claimed.

Given explicit formula: $$s_n = 5n - 2$$, where $$n = 1, 2, 3, \ldots$$

We have to write the first six terms of the sequence and then find the 100^th and 1000^th terms.

Step 1: First six terms

1. For $$n = 1$$:
   $$s_1 = 5(1) - 2 = 5 - 2 = 3$$

2. For $$n = 2$$:
   $$s_2 = 5(2) - 2 = 10 - 2 = 8$$

3. For $$n = 3$$:
   $$s_3 = 5(3) - 2 = 15 - 2 = 13$$

4. For $$n = 4$$:
   $$s_4 = 5(4) - 2 = 20 - 2 = 18$$

5. For $$n = 5$$:
   $$s_5 = 5(5) - 2 = 25 - 2 = 23$$

6. For $$n = 6$$:
   $$s_6 = 5(6) - 2 = 30 - 2 = 28$$

Thus, the first six terms are

Step 2: 100^th term

Substitute $$n = 100$$ in the formula:

$$s_{100} = 5(100) - 2 = 500 - 2 = 498$$

Step 3: 1000^th term

Substitute $$n = 1000$$ in the formula:

$$s_{1000} = 5(1000) - 2 = 5000 - 2 = 4998$$

Conclusion

• First six terms: $$3, 8, 13, 18, 23, 28$$
• 100^th term: $$498$$
• 1000^th term: $$4998$$

We are given a sequence defined recursively as follows:

Initial (first) term:
$$u_1 = 1$$

Recursive rule (for every integer $$n \ge 2$$):
$$u_n = 2u_{n-1} + 3$$

Step 1: Find the first four terms.

1. First term (already given): $$u_1 = 1$$
2. Second term:
   $$u_2 = 2u_{1} + 3 = 2\times1 + 3 = 2 + 3 = 5$$
3. Third term:
   $$u_3 = 2u_{2} + 3 = 2\times5 + 3 = 10 + 3 = 13$$
4. Fourth term:
   $$u_4 = 2u_{3} + 3 = 2\times13 + 3 = 26 + 3 = 29$$

Thus the first four terms are
$$1,\;5,\;13,\;29.$$

Step 2: Decide whether 133 is a term of the sequence.

Method 1: Obtain a general (explicit) formula and test 133.

The given recurrence is first-order linear:

$$u_n - 2u_{n-1}=3.$$

(i) Solve the homogeneous part $$u_n^{(h)}-2u_{n-1}^{(h)}=0$$.
Putting $$u_n^{(h)}=k^n$$ gives $$k^n-2k^{n-1}=0\Rightarrow k=2$$, so

$$u_n^{(h)} = A\,2^{n-1},$$ where $$A$$ is a constant.

(ii) Find one particular solution. Try a constant value $$u_n^{(p)} = C$$:

$$C - 2C = 3 \;\Longrightarrow\; -C = 3 \;\Longrightarrow\; C = -3.$$

(iii) Combine homogeneous and particular solutions:

$$u_n = A\,2^{n-1} - 3.$$

(iv) Use the initial condition $$u_1 = 1$$ to find $$A$$:

For $$n = 1$$, $$u_1 = A\,2^{1-1} - 3 = A - 3 = 1\;\Longrightarrow\; A = 4.$$

Hence the explicit (closed-form) formula is

\[u_n = 4\,2^{n-1} - 3.\]

Now test 133:

Set $$u_n = 133$$:

$$4\,2^{n-1} - 3 = 133$$
$$4\,2^{n-1} = 136$$
$$2^{n-1} = 34.$$

The number 34 is not a power of 2 (powers of 2 near it are 32 and 64). Therefore there is no integer $$n$$ for which $$2^{n-1}=34$$, so 133 is not a term of the sequence.

Method 2 (quick check by continuation):

The value 133 never appears; the terms jump from 125 to 253, so 133 is skipped. This agrees with the algebraic conclusion.

Conclusion

• The first four terms are $$1,\;5,\;13,\;29.$$
• Since $$2^{n-1} = 34$$ has no integer solution, 133 is not a term of the sequence.

The rule is recursive: each new term is calculated from the term that comes just before it.

Given

$$s_1 = 3,$$

and for every integer $$n \geq 2$$,

$$s_n = s_{n-1}\,(s_{n-1} - 1).$$

We need the first four terms: $$s_1, s_2, s_3, s_4$$.

1. First term (already given):
   $$s_1 = 3.$$

2. Second term (put $$n = 2$$ in the rule):

   $$s_2 = s_{1}\,(s_{1} - 1).$$

   Substitute $$s_1 = 3$$:

   $$s_2 = 3\,(3 - 1) = 3 \times 2 = 6.$$

3. Third term (put $$n = 3$$):

   $$s_3 = s_{2}\,(s_{2} - 1).$$

   Substitute $$s_2 = 6$$:

   $$s_3 = 6\,(6 - 1) = 6 \times 5 = 30.$$

4. Fourth term (put $$n = 4$$):

   $$s_4 = s_{3}\,(s_{3} - 1).$$

   Substitute $$s_3 = 30$$:

   $$s_4 = 30\,(30 - 1) = 30 \times 29 = 870.$$

Therefore, the first four terms of the sequence are $$3,\;6,\;30,\;870.$$

The taxi fare has two parts:

• a fixed booking fee of ₹200, which is paid once irrespective of the distance, and
• a running charge of ₹40 for each kilometre travelled.

Let the distance travelled be denoted by the positive integer $$n$$ (measured in km).

For a given $$n$$ kilometres, the total fare $$T_n$$ is therefore

We now list the fares for the first few natural numbers $$n = 1,2,3,\ldots$$:

The sequence of fares is therefore

Each term exceeds the previous one by a constant amount $$d = 40$$, so the sequence is an arithmetic progression (A.P.) with

• first term $$a = 240$$,
• common difference $$d = 40$$.

For an A.P., the nth term is given by $$T_n = a + (n-1)d$$. Substituting $$a = 240$$ and $$d = 40$$, we get back the earlier linear expression

Fare for 10 km

Put $$n = 10$$ in $$T_n = 200 + 40n$$:

The passenger must therefore pay ₹600 for a 10 km ride.

We recall the definition first.

Definition of a geometric progression (G.P.)
A sequence $$a_1, a_2, a_3,\dots$$ is called a geometric progression if the ratio of any term to its preceding term is the same constant, called the common ratio $$r$$; that is, $$\dfrac{a_{n+1}}{a_n}=r$$ for every admissible $$n$$.

Let us examine the given sequence:

$$a_1=1,\;a_2=2,\;a_3=4,\;a_4=8,\;a_5=16,\;\dots$$

1. Check the ratio of the 2nd term to the 1st term:
   $$\dfrac{a_2}{a_1}=\dfrac{2}{1}=2$$
2. Check the ratio of the 3rd term to the 2nd term:
   $$\dfrac{a_3}{a_2}=\dfrac{4}{2}=2$$
3. Check the ratio of the 4th term to the 3rd term:
   $$\dfrac{a_4}{a_3}=\dfrac{8}{4}=2$$
4. Check the ratio of the 5th term to the 4th term:
   $$\dfrac{a_5}{a_4}=\dfrac{16}{8}=2$$

Every time we divide a term by the preceding term, the result is the same number $$2$$. Since this ratio is constant, the sequence satisfies the definition of a geometric progression.

Therefore, the given sequence is a G.P., and its common ratio is

To decide whether the sequence 1, 3, 9, 27, 81, … is a geometric progression (G.P.), compare the ratio of each term to its preceding term.

For any sequence $$a_1, a_2, a_3, \ldots$$ to be a G.P., the following must be true:

$$\text{Common ratio } r \\ = \frac{a_{n+1}}{a_n}\; \text{is the same for all } n \ge 1.$$

Compute successive ratios for the given sequence:

First ratio:
$$r_1 = \frac{a_2}{a_1} = \frac{3}{1} = 3$$

Second ratio:
$$r_2 = \frac{a_3}{a_2} = \frac{9}{3} = 3$$

Third ratio:
$$r_3 = \frac{a_4}{a_3} = \frac{27}{9} = 3$$

Fourth ratio:
$$r_4 = \frac{a_5}{a_4} = \frac{81}{27} = 3$$

All calculated ratios are equal: $$r_1 = r_2 = r_3 = r_4 = 3$$.

Since the ratio $$\frac{a_{n+1}}{a_n}$$ remains constant and equals 3 for every pair of consecutive terms, the sequence is a geometric progression with common ratio 3.

Step 1 – Recall the definition of a geometric progression (G.P.)

A sequence $$a_1, a_2, a_3,\dots$$ is called a geometric progression if the ratio of any term (except the first) to the term just before it is a constant. That is, if there exists a non-zero number $$r$$ such that

$$\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=\dots=r$$,

then the sequence is a G.P. with common ratio $$r$$.

Step 2 – Write the first few terms of the given sequence

The sequence given is:

$$1,\; -1,\; 1,\; -1,\; 1,\; \dots$$

Thus

• $$a_1 = 1$$
• $$a_2 = -1$$
• $$a_3 = 1$$
• $$a_4 = -1$$

Step 3 – Find the ratio of consecutive terms

First pair:

$$\dfrac{a_2}{a_1}= \dfrac{-1}{1} = -1$$

Second pair:

$$\dfrac{a_3}{a_2}= \dfrac{1}{-1} = -1$$

Third pair:

$$\dfrac{a_4}{a_3}= \dfrac{-1}{1} = -1$$

We see that every time the ratio equals the same constant $$-1$$.

Step 4 – Conclusion

Because the ratio between any term and the term preceding it is the same constant $$-1$$, the sequence satisfies the definition of a geometric progression.

Hence,

Therefore, 1, –1, 1, –1, 1, … is indeed a geometric progression with common ratio $$-1$$.

Given sequence: $$5, \frac{15}{4}, \frac{45}{16}, \frac{135}{64}, \ldots$$

For a sequence to be a geometric progression (G.P.), the common ratio $$r$$ between any term and its preceding term must be constant:

$$r = \frac{a_{k+1}}{a_k}\; \text{for all} \; k \ge 1$$

We calculate $$r$$ for successive pairs:

1. Between the first and second terms:
   $$r = \frac{\frac{15}{4}}{5} = \frac{15}{4} \times \frac{1}{5} = \frac{15}{20} = \frac{3}{4}$$
2. Between the second and third terms:
   $$r = \frac{\frac{45}{16}}{\frac{15}{4}} = \frac{45}{16} \times \frac{4}{15} = \frac{180}{240} = \frac{3}{4}$$
3. Between the third and fourth terms:
   $$r = \frac{\frac{135}{64}}{\frac{45}{16}} = \frac{135}{64} \times \frac{16}{45} = \frac{2160}{2880} = \frac{3}{4}$$

Since the ratio is the same ($$r = \frac{3}{4}$$) for every pair of consecutive terms, the sequence is a geometric progression.

For a G.P. whose first term is $$a_1$$ and common ratio is $$r$$, the general (or $$n^{\text{th}}$$) term is

Here, $$a_1 = 5$$ and $$r = \frac{3}{4}$$, hence

This gives every term of the progression. For example, putting $$n = 1,2,3,\ldots$$ reproduces the given numbers.

Objective : For every sequence listed below we must

• observe a recurrent rule (pattern),
• justify that rule by showing intermediate calculations a Class 9 student is expected to carry out,
• write a usable algebraic description (the n^th or general term), and
• use the rule to generate the next three terms.

The sequences that were given just above the question are taken to be

1. $$1, 3, 5, 7, \dots$$

2. $$1, 4, 9, 16, \dots$$

3. $$2, 4, 8, 16, \dots$$

4. $$1, 2, 4, 7, 11, \dots$$

5. $$1, 1, 2, 3, 5, 8, \dots$$

Sequence (i) : $$1,3,5,7,\dots$$

• First–order differences: $$3-1 = 2,\;5-3 = 2,\;7-5 = 2$$. The difference is constantly $$2$$, so the sequence is an arithmetic progression (AP) with common difference $$d=2$$.

• General term of an AP: $$a_n = a_1 + (n-1)d$$. Here $$a_1 = 1,\;d = 2$$.

$$a_n = 1 + (n-1)\times 2 = 2n-1.$$

• Next three terms: put $$n = 5,6,7$$  ⇒  $$a_5=9,\;a_6=11,\;a_7=13$$.

Sequence (ii) : $$1,4,9,16,\dots$$

• These numbers are all perfect squares:

$$1 = 1^2,\;4 = 2^2,\;9 = 3^2,\;16 = 4^2.$$

• Therefore $$a_n = n^2$$ for $$n\ge 1$$.

• Next three terms correspond to $$n = 5,6,7$$:

$$a_5 = 5^2 = 25,\;a_6 = 6^2 = 36,\;a_7 = 7^2 = 49.$$

Sequence (iii) : $$2,4,8,16,\dots$$

• Each term is obtained by multiplying the preceding term by $$2$$ (common ratio $$r=2$$). The sequence is a geometric progression (GP).

• General term of a GP: $$a_n = a_1 r^{\,(n-1)}$$. Here $$a_1=2,\;r=2$$.

$$a_n = 2\times 2^{\,(n-1)} = 2^n.$$

• Next three terms ( $$n = 5,6,7$$ ):

$$a_5 = 2^5 = 32,\;a_6 = 2^6 = 64,\;a_7 = 2^7 = 128.$$

Sequence (iv) : $$1,2,4,7,11,\dots$$

• First–order differences: $$2-1 = 1,\;4-2 = 2,\;7-4 = 3,\;11-7 = 4$$ – the increments themselves form the sequence $$1,2,3,4,\dots$$ of successive natural numbers.

• Hence each term is obtained by adding, in order, $$1,2,3,4,\dots$$ to the previous term. Equivalently,

$$a_n = 1 + \underbrace{(1+2+3+\dots+(n-1))}_{\text{sum of first }(n-1)\text{ natural numbers}}.$$

The well-known formula $$1+2+\dots+(n-1)=\dfrac{(n-1)n}{2}$$ gives

$$a_n = 1 + \dfrac{(n-1)n}{2}.$$

• Next three terms: use $$n = 6,7,8$$.

$$\begin{aligned} a_6 &= 1 + \dfrac{5\times 6}{2} = 1+15 = 16,\\ a_7 &= 1 + \dfrac{6\times 7}{2} = 1+21 = 22,\\ a_8 &= 1 + \dfrac{7\times 8}{2} = 1+28 = 29. \end{aligned}$$

Sequence (v) : $$1,1,2,3,5,8,\dots$$

• Each term (starting from the third) is the sum of the two immediately preceding terms:

$$1+1=2,\;1+2=3,\;2+3=5,\;3+5=8,$$

and so on. This is the famous Fibonacci sequence.

• Using the recurrence $$a_{n}=a_{n-1}+a_{n-2}$$ with initial values $$a_1=a_2=1$$, find the next three terms:

$$\begin{aligned} a_7 &= a_5 + a_6 = 5 + 8 = 13,\\ a_8 &= a_6 + a_7 = 8 + 13 = 21,\\ a_9 &= a_7 + a_8 = 13 + 21 = 34. \end{aligned}$$

Summary of predictions

Thus, by recognising the underlying rule in each list, we can confidently extend every sequence.

Step 1 · Identify the pattern of the given sequence

The terms 1, 4, 7, 10, 13, … increase by the same amount each time.

Common difference:
$$d = 4 - 1 = 3 = 7 - 4 = 10 - 7 = \dots$$

So the sequence is an arithmetic progression (A.P.) with first term $$a = 1$$ and common difference $$d = 3$$.

Step 2 · Write the formula for the n-th term

For an A.P. the n-th term $$a_n$$ is

$$a_n = a + (n-1)d$$

Hence

$$a_n = 1 + (n-1)\,3 = 3n - 2$$

Step 3 · Predict the next four terms

Last known term is the 5th term, $$a_5 = 13$$.

• 6th term: $$a_6 = 3(6)-2 = 16$$
• 7th term: $$a_7 = 3(7)-2 = 19$$
• 8th term: $$a_8 = 3(8)-2 = 22$$
• 9th term: $$a_9 = 3(9)-2 = 25$$

Thus the next four terms are 16, 19, 22 and 25.

Step 4 · Define the required ‘running–sum’ sequence

Let $$S_n$$ be the sum of the first n terms of the original A.P. The question asks for the first ten values of $$S_n$$.

Step 5 · Derive a general formula for $$S_n$$

For an A.P. the sum of the first n terms is

$$S_n = \dfrac{n}{2}\,[2a + (n-1)d]$$

Substitute $$a = 1$$ and $$d = 3$$:

$$S_n = \dfrac{n}{2}\,[2(1) + (n-1)3]$$
$$\;\;\;\; = \dfrac{n}{2}\,[2 + 3n - 3]$$
$$\;\;\;\; = \dfrac{n}{2}\,[3n - 1]$$

Therefore

\[S_n = \dfrac{n\,(3n - 1)}{2}\]

Step 6 · Compute $$S_1, S_2, \dots , S_{10}$$

Step 7 · Write the running–sum sequence explicitly

The required sequence is:
1, 5, 12, 22, 35, 51, 70, 92, 117, 145.

All steps shown.

The triangular numbers arise from adding the first few natural numbers one after another.

In symbols, if $$t_n$$ denotes the $$n^{\text{th}}$$ triangular number, then

Using the well-known formula for the sum of the first $$n$$ natural numbers, we have

$$t_n = \dfrac{n(n+1)}{2}$$

Now substitute $$n = 5,6,7,8$$ one by one.

• For $$n = 5$$:
  $$t_5 = \dfrac{5(5+1)}{2} = \dfrac{5 \times 6}{2} = 15$$
• For $$n = 6$$:
  $$t_6 = \dfrac{6(6+1)}{2} = \dfrac{6 \times 7}{2} = 21$$
• For $$n = 7$$:
  $$t_7 = \dfrac{7(7+1)}{2} = \dfrac{7 \times 8}{2} = 28$$
• For $$n = 8$$:
  $$t_8 = \dfrac{8(8+1)}{2} = \dfrac{8 \times 9}{2} = 36$$

Thus the required terms are 15, 21, 28 and 36.

What is meant by a sequence?
In Class 9 we call an ordered list of numbers a sequence. If the numbers are written as $$a_1, a_2, a_3, \ldots , a_n, \ldots$$ then the sub-script tells us the position (1^st, 2^nd, 3^rd, …).

The textbook introduces arithmetic and geometric progressions. Here are five other kinds of sequences that you and your friends can explore.

1. Arithmetic Progression (AP)
   • Definition : the difference $$d$$ between consecutive terms is constant, i.e. $$a_{n+1}-a_n=d$$.
   • The n-th term is \[a_n=a_1+(n-1)d\]
   • Example : 2, 5, 8, 11, … has $$a_1=2$$, $$d=3$$.
     10^th term : $$a_{10}=2+(10-1)\times3=29$$.
   • Sum of first $$n$$ terms (proved in the text) \[S_n=\frac{n}{2}\,[2a_1+(n-1)d] \]
2. Geometric Progression (GP)
   • Definition : the ratio $$r$$ of consecutive terms is constant, i.e. $$\dfrac{a_{n+1}}{a_n}=r$$.
   • The n-th term is \[a_n=a_1 r^{\;n-1}\]
   • Example : 3, 6, 12, 24, … has $$a_1=3$$, $$r=2$$.
     6^th term : $$a_6=3\times2^{5}=96$$.
   • Sum of first $$n$$ terms (for $$r\neq1$$) \[S_n=a_1\,\frac{r^{\;n}-1}{r-1}\]
3. Fibonacci Sequence
   • Starts with $$F_1=1,\;F_2=1$$ and follows the recurrence relation \[F_n=F_{n-1}+F_{n-2}\quad(n\ge3)\]
   • Beginning terms : 1, 1, 2, 3, 5, 8, 13, …
   • Illustration : to find $$F_8$$ —
     $$F_3=1+1=2$$, $$F_4=2+1=3$$, $$F_5=3+2=5$$, $$F_6=5+3=8$$, $$F_7=8+5=13$$, $$F_8=13+8=21$$.
   • Property : the ratio $$\dfrac{F_{n+1}}{F_n}$$ approaches the golden ratio $$\varphi\,(\approx1.618)$$ as $$n$$ grows.
4. Square-number Sequence
   • Each term is the square of its position : $$a_n=n^2$$.
   • Terms : 1, 4, 9, 16, 25, …
   • 12^th term : $$a_{12}=12^2=144$$.
   • Property : successive differences form the odd numbers 3, 5, 7, 9, …
5. Triangular-number Sequence
   • Represents dots that can be arranged as an equilateral triangle.
   • Formula : $$T_n=\dfrac{n(n+1)}{2}$$.
   • Terms : 1, 3, 6, 10, 15, …
   • 10^th term : $$T_{10}=\dfrac{10\times11}{2}=55$$.
   • Property : differences between successive terms are the natural numbers 2, 3, 4, 5, …

Summary
These five sequences illustrate different rules: constant addition (AP), constant multiplication (GP), addition of two previous terms (Fibonacci), squaring the position (square numbers) and a quadratic formula giving triangular arrangements. Try plotting their terms or finding patterns in their differences and ratios for further exploration!

Understanding the idea of an explicit $n^{\text{th}}$-term formula

Suppose you are given a sequence $$a_1,\;a_2,\;a_3,\;\ldots$$. An explicit (or general) formula is a rule that produces the term $$a_n$$ directly from the position $$n$$:

\[a_n = f(n)\]

where $$f(n)$$ is some expression in $$n$$ (it could be linear, quadratic, a fraction, a power, … — whatever the sequence demands).

Why is such a formula useful?

1. Immediate calculation of any term
   Without the explicit rule you might have to write out all the preceding terms to reach, say, the $$50^{\text{th}}$$ term. With the rule you substitute once.
   Example. In the arithmetic progression $$2,5,8,11,\ldots$$ we have the explicit rule
   $$a_n = 2 + (n-1)\times3$$.
   To find the $$50^{\text{th}}$$ term:
   $$a_{50} = 2 + (50-1)\times3 = 2 + 49\times3 = 2 + 147 = 149.$$
   No listing of the first 49 terms is required.
2. Testing whether a given number is in the sequence
   Because $$a_n$$ is expressed in $$n$$ you can turn the question “Is $$k$$ a term?” into an algebraic equation and see whether it gives an integer $$n$$.
   Example. Using the same rule $$a_n = 2 + 3(n-1)$$, check if $$203$$ is a term:
   
   • Set $$2 + 3(n-1) = 203.$$
   • Solve: $$3(n-1) = 201 \;\Rightarrow\; n-1 = 67 \;\Rightarrow\; n = 68.$$
   • Because $$n = 68$$ is a positive integer, $$203$$ is the $$68^{\text{th}}$$ term.
3. Proving properties of the sequence
   Many theoretical results (for example, verifying that the difference between consecutive terms is constant, or establishing divisibility patterns) are far easier when you can manipulate the general term algebraically.
4. Finding other formulas such as the sum of the first $n$ terms
   In an arithmetic progression the derivation of the sum formula
   \[S_n = \dfrac{n}{2}\,(a_1 + a_n)\] hinges on first having $$a_n$$ expressed explicitly.
5. Efficient use of technology
   A calculator or a computer spreadsheet can generate large parts of the sequence instantly once you supply the explicit rule. This is impractical if each term depends on all the previous ones.

Conclusion

A single compact formula $$a_n = f(n)$$ turns the sequence into something you can examine, compute, and prove facts about directly, without repeatedly building it term-by-term. That is why having an explicit $$n^{\text{th}}$$-term formula is so useful.

The sequence of odd natural numbers can be generated by the explicit ("nth-term") rule

$$u_n = 2n - 1,$$

where $$n$$ is the position (rank) of the term in the sequence and $$u_n$$ is the value of that term.

We must evaluate $$u_n$$ for three different values of $$n$$.

1. 53^rd term
   Substitute $$n = 53$$:
   $$u_{53} = 2 \times 53 - 1 = 106 - 1 = 105.$$
2. 108^th term
   Substitute $$n = 108$$:
   $$u_{108} = 2 \times 108 - 1 = 216 - 1 = 215.$$
3. 1170^th term
   Substitute $$n = 1170$$:
   $$u_{1170} = 2 \times 1170 - 1 = 2340 - 1 = 2339.$$

Hence the required terms are:

• 53^rd term = 105,
• 108^th term = 215,
• 1170^th term = 2339.

The sequence of square numbers begins

$$1,\;4,\;9,\;16,\;25,\;\ldots$$

Label the terms as $$t_1,\,t_2,\,t_3,\ldots$$ so that $$t_n$$ denotes the $$n^{\text{th}}$$ term we wish to describe.

Step 1 – Look at the first few terms

Notice that every term equals the number of its position squared:

• First term: $$1 = 1 \times 1 = 1^2$$
• Second term: $$4 = 2 \times 2 = 2^2$$
• Third term: $$9 = 3 \times 3 = 3^2$$
• Fourth term: $$16 = 4 \times 4 = 4^2$$
• Fifth term: $$25 = 5 \times 5 = 5^2$$

The pattern strongly suggests $$t_n = n^2$$, but we verify it algebraically.

Step 2 – Confirm by differences

Find successive differences of the sequence:

The differences are the consecutive odd numbers $$3,5,7,9,\ldots$$. The sum of the first $$n$$ odd numbers is known to be $$n^2$$. Hence if $$t_n = n^2$$, each increase must indeed be an odd number, which matches the table, confirming the rule.

Step 3 – State the general term

Because every check is consistent, the $$n^{\text{th}}$$ term is

Therefore the required rule is $$t_n = n^2$$.

Step 1 : Observe and record the first few stages

The number of new squares that appear each time forms the sequence $$3,5,7,9,\dots$$ – the consecutive odd numbers.

Step 2 : Express the nth difference

The odd numbers can be written as

$$3 = 2 \times 2 - 1,\; 5 = 2 \times 3 - 1,\; 7 = 2 \times 4 - 1,\;\dots$$

In general, at Stage $$n \,(n \ge 2)$$ the extra squares added are

$$2n - 1.$$

Step 3 : Form a cumulative (recursive) relation

Let $$S_n$$ be the total number of squares in Stage $$n$$. Then

$$S_1 = 1,\qquad S_{n} = S_{n-1} + (2n-1) \;\;\text{for } n \ge 2.$$

Step 4 : Solve the recurrence to get a direct formula

Add the first $$n$$ odd numbers:

$$S_n = 1 + 3 + 5 + \dots + (2n-1).$$

A known result (which can be proved by induction) is

\[S_n = n^2.\]

Thus Stage $$n$$ contains exactly $$n^2$$ small squares.

Step 5 : Predict the requested stages

• Stage 5: $$S_5 = 5^2 = 25$$ squares.
• Stage 6: $$S_6 = 6^2 = 36$$ squares.
• Stage 10: $$S_{10} = 10^2 = 100$$ squares.
• Stage 11: $$S_{11} = 11^2 = 121$$ squares.
• Stage 12: $$S_{12} = 12^2 = 144$$ squares.
• Stage 20: $$S_{20} = 20^2 = 400$$ squares.

Step 6 : General prediction

At any stage $$n$$ the number of squares is $$n^2.$$

Recall – definition of an arithmetic progression (A.P.)
A sequence $$a_1 , a_2 , a_3 ,\ldots$$ is an A.P. iff the difference between every two consecutive terms is the same; that is,

The constant number $$d$$ is called the common difference.

The sequences that have appeared in the chapter are the following:

1. $$1,2,3,4,5,\ldots$$
2. $$2,4,6,8,10,\ldots$$
3. $$5,10,15,20,25,\ldots$$
4. $$1,4,9,16,25,\ldots$$ (perfect squares)
5. $$1,3,5,7,9,\ldots$$
6. $$1,2,4,8,16,\ldots$$ (powers of 2)

We check the common difference for each sequence.

Final classification

• Arithmetic progressions: Sequences (i), (ii), (iii) and (v).
• Not arithmetic progressions: Sequences (iv) and (vi).

This follows directly from the fact that only in the first, second, third and fifth sequences is the difference between successive terms the same throughout; in the fourth and sixth sequences the difference keeps changing, so the defining condition of an A.P. fails.

We are given the general term (or nth term) of a sequence:

$$t_n = 5n - 3 \quad (n \ge 1).$$

To obtain any particular term, substitute the required value of $$n$$ into the expression.

1. 10^th term

Take $$n = 10$$.

\[t_{10} = 5(10) - 3\]

First multiply: $$5 \times 10 = 50$$.

Then subtract 3: $$50 - 3 = 47$$.

Therefore, $$t_{10} = 47$$.

2. 15^th term

Take $$n = 15$$.

\[t_{15} = 5(15) - 3\]

First multiply: $$5 \times 15 = 75$$.

Then subtract 3: $$75 - 3 = 72$$.

Therefore, $$t_{15} = 72$$.

Hence, the 10^th term is $$47$$ and the 15^th term is $$72$$.

The sequence is defined by its general (nth) term

$$t_n = 5n - 3, \; n \ge 1$$

To check whether a given number belongs to the sequence, equate it with $$t_n$$ and solve for $$n$$. The number will be a term of the sequence only if the resulting $$n$$ is a positive integer.

1. Checking whether 97 is a term

Set $$t_n = 97$$:

$$5n - 3 = 97$$

Add 3 to both sides:

$$5n = 97 + 3 = 100$$

Divide by 5:

$$n = \dfrac{100}{5} = 20$$

Since $$n = 20$$ is a positive integer (and $$20 \ge 1$$), 97 is indeed the 20th term, i.e. $$97 = t_{20}$$.

2. Checking whether 172 is a term

Set $$t_n = 172$$:

$$5n - 3 = 172$$

Add 3 to both sides:

$$5n = 172 + 3 = 175$$

Divide by 5:

$$n = \dfrac{175}{5} = 35$$

Since $$n = 35$$ is a positive integer (and $$35 \ge 1$$), 172 is the 35th term, i.e. $$172 = t_{35}$$.

Conclusion

Both 97 and 172 correspond to positive integer values of $$n$$ (20 and 35, respectively). Therefore both numbers are terms of the given sequence.

The nth (general) term of the given arithmetic sequence is

$$t_n = 5n - 3, \; n \ge 1$$

We are asked to find which term equals 607, that is, to find the natural number $$n$$ such that

\[t_n = 607\]

Substituting the expression for $$t_n$$:

$$5n - 3 = 607$$

Now solve this simple linear equation step by step.

1. Add 3 to both sides:

$$5n - 3 + 3 = 607 + 3$$
$$5n = 610$$

2. Divide both sides by 5:

$$n = \dfrac{610}{5}$$
$$n = 122$$

Because $$n = 122$$ is a positive integer, it fits the condition $$n \ge 1$$. Hence,

\[\textbf{The 122nd term of the sequence is 607.}\]

Step 1 : Identify the type of sequence
Because every term is obtained by adding the fixed number $$3$$ to the preceding term, the sequence is an arithmetic progression (A.P.).
For an A.P. we have

• first term $$a = t_1$$,
• common difference $$d = t_{n+1}-t_n$$.

Step 2 : Read off the first term and the common difference
Given $$t_1 = -5$$, so $$a = -5$$.
Given $$t_{n+1} = t_n + 3$$, so each time we add $$3$$. Hence the common difference is
$$d = 3$$.

Step 3 : Write the first five terms
Start with $$t_1$$ and keep adding $$d = 3$$.

• First term : $$t_1 = -5$$
• Second term : $$t_2 = t_1 + 3 = -5 + 3 = -2$$
• Third term : $$t_3 = t_2 + 3 = -2 + 3 = 1$$
• Fourth term : $$t_4 = t_3 + 3 = 1 + 3 = 4$$
• Fifth term : $$t_5 = t_4 + 3 = 4 + 3 = 7$$

Therefore the first five terms are
$$-5,\\ -2,\ 1,\ 4,\ 7.$$

Step 4 : Write the general (nth) term
For an A.P. the general term is
\[ t_n = a + (n-1)d \] Substituting $$a = -5$$ and $$d = 3$$, we get
$$t_n = -5 + (n-1)\times 3.$$

Step 5 : Check whether 52 is a term
We must see if some positive integer $$n$$ satisfies
$$t_n = 52.$$

Set the expression for $$t_n$$ equal to 52:

Solve for $$n$$ step by step:

• Add 5 to both sides:
  $$ (n-1)\times 3 = 52 + 5 = 57 $$
• Divide both sides by 3:
  $$ n - 1 = \frac{57}{3} = 19 $$
• Add 1 to both sides:
  $$ n = 19 + 1 = 20 $$

$$n = 20$$ is a positive integer, so 52 is indeed a term of the sequence, namely the 20^th term.

Conclusion
First five terms: $$-5,\;-2,\;1,\;4,\;7.$$
Yes, 52 appears and it is the 20^th term.

The sequence is defined by

$$T_1 = 1, \; T_2 = 2, \; T_3 = 4$$

and for every integer $$n \ge 4$$,

$$T_n = T_{n-1} + T_{n-2} + T_{n-3}.$$

We calculate each required term one by one.

Term $$T_4$$

$$\begin{aligned} T_4 &= T_3 + T_2 + T_1\\ &= 4 + 2 + 1\\ &= 7 \end{aligned}$$

Term $$T_5$$

$$\begin{aligned} T_5 &= T_4 + T_3 + T_2\\ &= 7 + 4 + 2\\ &= 13 \end{aligned}$$

Term $$T_6$$

$$\begin{aligned} T_6 &= T_5 + T_4 + T_3\\ &= 13 + 7 + 4\\ &= 24 \end{aligned}$$

Term $$T_7$$

$$\begin{aligned} T_7 &= T_6 + T_5 + T_4\\ &= 24 + 13 + 7\\ &= 44 \end{aligned}$$

Term $$T_8$$

$$\begin{aligned} T_8 &= T_7 + T_6 + T_5\\ &= 44 + 24 + 13\\ &= 81 \end{aligned}$$

Thus the next five terms of the sequence are:

\[T_4 = 7, \; T_5 = 13, \; T_6 = 24, \; T_7 = 44, \; T_8 = 81\]

The given numbers 3, 8, 13, 18, \(\dots\) form an arithmetic progression (AP) because the difference between consecutive terms is constant.

First term:
$$a_1 = 3$$

Common difference:
$$d = 8 - 3 = 5$$
(You can verify that $$13 - 8 = 5$$ and $$18 - 13 = 5$$ as well.)

The $$n^{\text{th}}$$ term $$a_n$$ of an AP is found from

(i) 10^th term

Substitute $$n = 10$$:

$$a_{10} = 3 + (10 - 1) \times 5$$

$$a_{10} = 3 + 9 \times 5$$

$$a_{10} = 3 + 45 = 48$$

(ii) 26^th term

Substitute $$n = 26$$:

$$a_{26} = 3 + (26 - 1) \times 5$$

$$a_{26} = 3 + 25 \times 5$$

$$a_{26} = 3 + 125 = 128$$

Hence, the 10^th term is $$48$$ and the 26^th term is $$128$$.

First identify the parameters of the arithmetic progression (AP):

• First term: $$a = 21$$
• Common difference: $$d = 18 - 21 = -3$$

The general (n^th) term of an AP is

\[a_n = a + (n-1)d\]

(i) Finding the term number whose value is –81

Put $$a_n = -81$$ and solve for $$n$$:

$$-81 = 21 + (n-1)(-3)$$

$$-81 - 21 = (n-1)(-3)$$

$$-102 = (n-1)(-3)$$

$$n - 1 = \frac{-102}{-3} = 34$$

$$n = 34 + 1 = 35$$

Hence, –81 is the 35^th term of the AP.

(ii) Checking whether 0 is a term of the AP

Put $$a_n = 0$$ and solve for $$n$$:

$$0 = 21 + (n-1)(-3)$$

$$-21 = (n-1)(-3)$$

$$n - 1 = \frac{-21}{-3} = 7$$

$$n = 7 + 1 = 8$$

The value of $$n$$ obtained is a positive integer (8). Therefore 0 is a term of the given AP, specifically the 8^th term.

Step 1 • Identify first term and common difference

The given arithmetic progression (AP) is

$$11,\;8,\;5,\;2,\;\ldots$$

The first term is

$$a_1 = 11$$

The common difference $$d$$ is obtained by subtracting any term from the next term:

$$d = 8 - 11 = -3$$

We may check with another pair to be sure:

$$5 - 8 = -3,\;\;2 - 5 = -3 \;\;(\text{same } d).$$

Step 2 • General (\(n^{\text{th}}\)) term

For any AP, the general term $$a_n$$ is given by

$$a_n = a_1 + (n-1)d$$

Substitute $$a_1 = 11$$ and $$d = -3$$:

$$a_n = 11 + (n-1)(-3)$$

Simplify step-by-step:

$$a_n = 11 - 3(n-1)$$

Distribute the $$-3$$:

$$a_n = 11 - 3n + 3$$

Combine like terms $$11 + 3 = 14$$:

$$a_n = 14 - 3n$$

The required \(n^{\text{th}}\) term is therefore

Step 3 • Recursive rule

To express the AP recursively we start with the first term and relate each term to the previous one.

• Initial condition: $$a_1 = 11$$
• Recursive relation: each term is obtained by subtracting 3 from the previous term, i.e.

  $$a_{n+1} = a_n - 3\;\;\text{for } n \ge 1$$

Conclusion

The AP has \(n^{\text{th}}\) term $$a_n = 14 - 3n$$ and the recursive rule

$$a_1 = 11,\;a_{n+1} = a_n - 3.$$

An arithmetic progression (AP) is completely fixed once we know its first term $$a$$ and common difference $$d$$. For every positive integer $$n$$, the formula for the $$n^{\text{th}}$$ term is

We are told:

• The 3^rd term is 12, so $$T_3=12$$.
• The AP has 50 terms and the last (i.e. 50^th) term is 106, so $$T_{50}=106$$.

Translate each fact with the general term formula.

For $$n=3:$$ $$T_3=a+(3-1)d=a+2d=12 \;\;\;(1)$$

For $$n=50:$$ $$T_{50}=a+(50-1)d=a+49d=106 \;\;\;(2)$$

Now solve the simultaneous linear equations (1) and (2).

Subtract (1) from (2):

$$\bigl(a+49d\bigr)-\bigl(a+2d\bigr)=106-12$$
$$a+49d-a-2d=94$$
$$47d=94$$
$$d=\dfrac{94}{47}=2$$

Substitute $$d=2$$ into equation (1):

$$a+2(2)=12 \;\Rightarrow\; a+4=12 \;\Rightarrow\; a=8$$

Thus, the AP is $$8,\;10,\;12,\;\dots$$ with $$d=2$$.

The 29^th term is

$$T_{29}=a+(29-1)d=a+28d$$

Insert $$a=8$$ and $$d=2$$:

$$T_{29}=8+28\times2=8+56=64$$

Therefore, the 29^th term of the AP is 64.

Step 1 : Identify the range of 2-digit numbers.

The two-digit natural numbers run from $$10$$ to $$99$$.

Step 2 : Find the smallest and largest 2-digit multiples of 3.

A number is divisible by $$3$$ when the remainder on division by $$3$$ is $$0$$. Checking the first few two-digit numbers, $$10$$ leaves remainder $$1$$, $$11$$ leaves remainder $$2$$, and $$12 = 3 \times 4$$ is exactly divisible. So the smallest 2-digit multiple of $$3$$ is $$12$$.

The largest 2-digit number is $$99$$, and $$99 \div 3 = 33$$ with no remainder, so the largest 2-digit multiple of $$3$$ is $$99$$.

Step 3 : Write the resulting sequence.

All the required numbers form the arithmetic progression

$$12,\; 15,\; 18,\; \ldots,\; 99$$

with first term $$a = 12$$ and common difference $$d = 3$$.

Step 4 : Find the number of terms $$n$$.

Using the formula for the $$n^{\text{th}}$$ term of an AP, $$l = a + (n-1)d$$:

\[99 = 12 + (n-1)\times 3\]

$$\Rightarrow\; 99 - 12 = (n-1)\times 3$$

$$\Rightarrow\; 87 = (n-1)\times 3$$

$$\Rightarrow\; n - 1 = 29$$

$$\Rightarrow\; n = 30.$$

Therefore, there are 30 two-digit numbers divisible by $$3$$.

Step 5 : Find the sum of these numbers.

The sum of all terms of an AP is

$$S = \dfrac{n}{2}\,(a + l).$$

Substituting $$n = 30$$, $$a = 12$$ and $$l = 99$$:

\[S = \dfrac{30}{2}\,(12 + 99) = 15 \times 111 = 1665.\]

So, the sum of all the 2-digit numbers divisible by $$3$$ is 1665.

Step 1 Identify the sequence of salaries
Harish’s salaries form an arithmetic progression (A.P.) because the amount increases by a fixed sum every year.

• First term (salary in the 1^st year): $$a = 5{,}00{,}000$$
• Common difference (annual increment): $$d = 20{,}000$$
• Let $$n$$ be the number of years after which the salary becomes $$7{,}00{,}000$$.

Step 2 Write the n-th term of the A.P.
For an arithmetic progression, the salary in the $$n$$-th year is given by

$$a_n = a + (n-1)d$$

Step 3 Set the n-th term equal to ₹7,00,000

Substituting the known values:

$$5{,}00{,}000 + (n-1)\times 20{,}000 = 7{,}00{,}000$$

Step 4 Solve for $$n$$

First isolate the term containing $$n$$:

$$ (n-1)\times 20{,}000 = 7{,}00{,}000 - 5{,}00{,}000 $$

$$ (n-1)\times 20{,}000 = 2{,}00{,}000 $$

Divide both sides by $$20{,}000$$:

$$ n - 1 = \frac{2{,}00{,}000}{20{,}000} = 10 $$

Add 1 to both sides:

$$ n = 10 + 1 = 11 $$

Step 5 Interpretation
The salary reaches ₹7,00,000 in the 11^th year.

The marbles in each row form the sequence

$$1, 2, 3, \ldots , 25$$

This is an arithmetic progression (AP) with common difference

$$d = 2 - 1 = 1$$

We need the sum of the first $$n = 25$$ terms of this AP.

For an AP, the sum $$S_n$$ of the first $$n$$ terms is

\[S_n = \frac{n}{2}(a + l)\]

where

• first term $$a = 1$$,
• last term $$l = 25$$,
• number of terms $$n = 25$$.

Substitute these values:

$$S_{25} = \frac{25}{2}(1 + 25)$$

$$\phantom{S_{25}} = \frac{25}{2} \times 26$$

$$\phantom{S_{25}} = 25 \times 13$$

$$\phantom{S_{25}} = 325$$

Hence, the total number of marbles used is

\[325\text{ marbles}.\]

The n^th term of a geometric progression (GP) with first term $$a$$ and common ratio $$r$$ is

$$a_n = a r^{n-1}.$$

Given the common ratio $$r = 2$$ and the 8^th term $$a_8 = 192$$, we substitute $$n = 8$$:

$$a_8 = a r^{8-1} = a (2)^{7}.$$

Hence,

$$a \times 128 = 192 \quad \Rightarrow \quad a = \frac{192}{128} = \frac{3}{2}.$$

Thus the first term is $$a = \frac{3}{2}$$.

To find the 12^th term:

$$a_{12} = a r^{12-1} = \frac{3}{2} \times 2^{11}.$$

Since $$2^{11} = 2048$$,

$$a_{12} = \frac{3}{2} \times 2048 = 3 \times 1024 = 3072.$$

Therefore,

\[a_{12} = 3072\]

Step 1 — Identify the first term and common ratio

The given sequence is 5, 25, 125, … .

• First term: $$a = 5$$
• Common ratio: $$r = \dfrac{25}{5} = 5$$ (and $$\dfrac{125}{25}=5$$, so the ratio is constant)

Therefore the sequence is a geometric progression (GP) with $$a = 5$$ and $$r = 5$$.

Step 2 — Formula for the $$n^{\text{th}}$$ term of a GP

For any GP, the $$n^{\text{th}}$$ term $$a_n$$ is given by

$$a_n = a r^{\,n-1}$$  for $$n \ge 1$$.

Step 3 — Find the 10^th term

Substitute $$n = 10$$ in the formula:

$$a_{10} = a r^{10-1} = 5 \times 5^{9}$$

Combine the bases (recall $$5 \times 5^{9} = 5^{1} \times 5^{9} = 5^{10}$$):

\[a_{10} = 5^{10}\]

Calculate the value step by step:

• $$5^2 = 25$$
• $$5^3 = 125$$
• $$5^4 = 625$$
• $$5^5 = 3\,125$$
• $$5^6 = 15\,625$$
• $$5^7 = 78\,125$$
• $$5^8 = 390\,625$$
• $$5^9 = 1\,953\,125$$
• $$5^{10} = 9\,765\,625$$

Hence the 10^th term is $$9\,765\,625$$.

Step 4 — Find the general ( $$n^{\text{th}}$$ ) term

Using $$a = 5$$ and $$r = 5$$ in the formula $$a_n = a r^{\,n-1}$$:

$$a_n = 5 \times 5^{\,n-1}$$

Combine the bases (add exponents):

$$a_n = 5^{1} \times 5^{\,n-1} = 5^{n}$$

\[\boxed{\;a_n = 5^{n}\;}\]

Conclusion

• 10^th term: $$a_{10} = 5^{10} = 9\,765\,625$$
• General term: $$a_n = 5^{n}$$

We are given the sequence defined recursively by

$$t_1 = 2, \; t_{n+1} = 3t_n - 2\;(n \ge 1).$$

The task is to locate the value of n for which $$t_n = 730$$.

Step 1: Find a closed-form (general) expression for $$t_n$$.

This is a linear first-order recurrence of the pattern $$t_{n+1} - 3t_n = -2$$.
We tackle it in two parts: a homogeneous solution and a particular solution.

• Homogeneous part: Set the right–hand side to 0.
     $$t_{n+1}^{(h)} - 3t_n^{(h)} = 0 \;\Rightarrow\; t_{n+1}^{(h)} = 3t_n^{(h)}.$$
     Thus $$t_n^{(h)} = A\,3^{\,n-1},$$ where $$A$$ is a constant.
• Particular part: Look for a constant solution $$t_n^{(p)} = C\;(\text{independent of }n).$$
     Substitute: $$C - 3C = -2 \;\Rightarrow\; -2C = -2 \;\Rightarrow\; C = 1.$$

Add the two parts:

\[t_n = A\,3^{\,n-1} + 1\]

Step 2: Determine the value of the constant $$A$$ using $$t_1 = 2$$.

$$t_1 = A\,3^{0} + 1 = A + 1 = 2 \;\Rightarrow\; A = 1.$$

Hence the explicit formula is

$$t_n = 3^{\,n-1} + 1.$$

Step 3: Solve $$t_n = 730$$.

Set $$t_n$$ equal to the required value:

$$3^{\,n-1} + 1 = 730 \;\Rightarrow\; 3^{\,n-1} = 729.$$

Recall that $$729 = 3^6.$

Therefore $$3^{\,n-1} = 3^6 \;\Rightarrow\; n - 1 = 6 \;\Rightarrow\; n = 7.$$

Step 4: Verification (optional but reassuring).

Using the closed form: $$t_7 = 3^{6} + 1 = 729 + 1 = 730,$$ which matches the target value.

Consequently, the number 730 appears as the 7^th term of the sequence.

Step 1  Identify the first term and the common ratio

The given GP is 2, 6, 18, …

• First term: $$a = 2$$
• Common ratio: $$r = \frac{6}{2} = 3$$ (and $$\frac{18}{6}=3$$, confirming the ratio).

Step 2  Write the general (explicit) formula for the n^th term

For any GP the explicit formula is $$a_n = a\,r^{\,n-1}$$. Substituting $$a=2$$ and $$r=3$$, we get

\[a_n = 2 \cdot 3^{\,n-1}\]

Step 3  Use the formula to locate the term equal to 4374

We set $$a_n = 4374$$ and solve for n:

$$2 \cdot 3^{\,n-1} = 4374$$

Divide both sides by 2:

$$3^{\,n-1} = \frac{4374}{2} = 2187$$

Step 4  Express 2187 as a power of 3

Note that

$$3^5 = 243, \; 3^6 = 729, \; 3^7 = 2187$$

Hence $$3^{\,n-1} = 3^7$$, which implies $$n-1 = 7$$.

Therefore $$n = 7 + 1 = 8$$.

Conclusion

The number 4374 is the 8^th term of the progression.

Explicit formula

\[a_n = 2 \cdot 3^{\,n-1}\]

Recursive formula

• Initial term: $$a_1 = 2$$
• Recursion: $$a_n = 3\,a_{n-1}\quad (n \ge 2)$$

The sequence is

$$2,\;2\sqrt{2},\;4,\dots$$

Step 1 – Identify the common ratio.

• First term: $$a = 2$$
• Second ÷ first: $$r = \dfrac{2\sqrt{2}}{2}=\sqrt{2}$$
• Third ÷ second: $$\dfrac{4}{2\sqrt{2}} = \sqrt{2}$$ (confirms the same ratio)

Hence it is a geometric progression (G.P.) with $$a = 2$$ and $$r = \sqrt{2}$$.

Step 2 – Write the n-th term.

For a G.P., $$T_n = a\,r^{\,n-1}$$.

Here, $$T_n = 2\,(\sqrt{2})^{n-1}$$.

Step 3 – Set the n-th term equal to 128.

$$2\,(\sqrt{2})^{n-1} = 128$$

Divide by 2:

$$(\sqrt{2})^{n-1} = 64$$

Step 4 – Express both sides with the same base.

$$64 = 2^6$$ and $$\bigl(\sqrt{2}\bigr)^{n-1} = \bigl(2^{1/2}\bigr)^{n-1} = 2^{\tfrac{n-1}{2}}$$

So

$$2^{\tfrac{n-1}{2}} = 2^6$$

Step 5 – Equate exponents.

$$\frac{n-1}{2} = 6 \;\Longrightarrow\; n - 1 = 12 \;\Longrightarrow\; n = 13$$

Therefore, 128 is the 13^th term of the given sequence.

Let the first term of the arithmetic progression be $$a$$ and the common difference be $$d$$.

For any AP the $$n^{\text{th}}$$ term is $$a_n = a + (n-1)d$$.

Step 1 – Set up equations from the given terms

11^th term:
$$a_{11}=a+(11-1)d=a+10d=38 \quad(1)$$

16^th term:
$$a_{16}=a+(16-1)d=a+15d=73 \quad(2)$$

Step 2 – Find d

Subtract (1) from (2):

$$\bigl(a+15d\bigr)-\bigl(a+10d\bigr)=73-38$$

$$5d=35$$

$$d=7$$

Step 3 – Find a

Substitute $$d=7$$ into (1):

$$a+10(7)=38$$

$$a+70=38$$

$$a=38-70=-32$$

Step 4 – Find the 31^st term

$$a_{31}=a+(31-1)d=a+30d$$

$$a_{31}=-32+30\times7$$

$$a_{31}=-32+210=178$$

Hence, the 31^st term of the AP is 178.

Given
The 3^rd term of an arithmetic progression (AP) is 16.
The 7^th term exceeds the 5^th term by 12.

Step 1 · Write the general term
For an AP the n^th term is $$T_n = a + (n-1)d$$ where $$a$$ is the first term and $$d$$ is the common difference.

Step 2 · Translate the first condition
3^rd term: $$T_3 = a + (3-1)d = a + 2d$$
We are told $$T_3 = 16$$, therefore
$$a + 2d = 16 \ \ \ (1)$$

Step 3 · Translate the second condition
7^th term: $$T_7 = a + (7-1)d = a + 6d$$
5^th term: $$T_5 = a + (5-1)d = a + 4d$$
“7^th term exceeds the 5^th term by 12” means
$$T_7 - T_5 = 12$$

Substitute the expressions:
$$\big(a + 6d\big) - \big(a + 4d\big) = 12$$
Simplify:
$$a + 6d - a - 4d = 12$$
$$2d = 12$$
Therefore
$$d = 6 \ \ \ (2)$$

Step 4 · Find the first term
Substitute $$d = 6$$ in equation (1):
$$a + 2(6) = 16$$
$$a + 12 = 16$$
$$a = 16 - 12 = 4$$

Step 5 · Write the AP
The required AP has first term $$a = 4$$ and common difference $$d = 6$$.
Thus the sequence is
4, 10, 16, 22, 28, …

Verification (optional)
3^rd term: $$4 + 2 \times 6 = 16$$ ✔️
Difference between 7^th and 5^th terms:
$$\big(4 + 6 \times 6\big) - \big(4 + 4 \times 6\big) = (4 + 36) - (4 + 24) = 40 - 28 = 12$$ ✔️

Therefore, the required AP is

Step 1 : Identify the first three-digit multiple of 7

The smallest three-digit number is $$100$$. Divide by $$7$$ to locate the next higher multiple:

$$100 \div 7 = 14 \text{ remainder } 2.$$

Since a remainder appears, move to the next integer quotient, $$14+1 = 15$$.

Hence the first three-digit multiple of 7 is

$$7 \times 15 = 105.$$

Step 2 : Identify the last three-digit multiple of 7

The largest three-digit number is $$999$$. Divide by $$7$$:

$$999 \div 7 = 142 \text{ remainder } 5.$$

The remainder shows that $$999$$ itself is not divisible by $$7$$. The previous integer quotient is $$142$$, so the corresponding multiple is

$$7 \times 142 = 994.$$

Therefore the greatest three-digit multiple of 7 is $$994$$.

Step 3 : Recognise the numbers form an A.P.

All three-digit multiples of 7 form the sequence

$$105,\;112,\;119,\;\ldots,\;994$$

which is an arithmetic progression (A.P.) with

• first term $$a = 105$$
• common difference $$d = 7$$
• last term (n^th term) $$l = 994$$.

Step 4 : Use the nth-term formula to find how many terms

The nth term of an A.P. is

$$a_n = a + (n-1)d.$$

Put $$a_n = 994,\; a = 105,\; d = 7$$:

$$994 = 105 + (n-1)\times 7.$$

Simplify:

$$994 - 105 = (n-1)\times 7$$

$$889 = (n-1)\times 7$$

$$n - 1 = \dfrac{889}{7} = 127$$

$$n = 127 + 1 = 128.$$

Step 5 : Conclusion

There are 128 three-digit numbers that are divisible by 7.

We are asked to count the multiples of 4 that are strictly between 10 and 250.

Step 1 ∙ Identify the first and last relevant multiples of 4

• The number immediately bigger than 10 that is divisible by 4 is
  $$10\div 4 = 2.5 \;\text{(remainder)} \;\Rightarrow\; 3\times4 = 12.$$
  So the smallest required multiple is $$a = 12.$$
• The greatest number below 250 that is divisible by 4 is obtained similarly:
  $$250\div 4 = 62.5 \;\text{(remainder)} \;\Rightarrow\; 62\times4 = 248.$$
  Hence the largest required multiple is $$l = 248.$$

Step 2 ∙ Recognise the arithmetic progression (AP)

All multiples of 4 constitute an AP with common difference $$d = 4.$$ Here the AP that interests us is

$$12,\;16,\;20,\;\dots,\;248.$$

Step 3 ∙ Use the nth-term formula to count terms

For an AP, the nth term is $$a_n = a + (n-1)d.$$ We know the last term $$a_n = l = 248$$, so

$$248 = 12 + (n-1)\times4.$$

Isolate $$n$$:

\[\begin{aligned} 248 - 12 &= (n-1)\times4\\[2pt] 236 &= 4(n-1)\\[2pt] \frac{236}{4} &= n-1\\[2pt] 59 &= n-1\\[2pt] n &= 60. \end{aligned}\]

Step 4 ∙ State the result

There are $$n = 60$$ multiples of 4 lying between 10 and 250.

Let the first term of the geometric progression (G.P.) be $$a$$ and the common ratio be $$r$$.

Then the terms of the G.P. are

$$a,\;ar,\;ar^2,\;ar^3,\;ar^4,\ldots$$

1. Translating the given conditions

1. The sum of the first two terms is –4: $$a+ar=-4$$ which simplifies to $$a(1+r)=-4.\quad(1)$$
2. The 5^th term is four times the 3^rd term: $$ar^4=4\,(ar^2)$$ Dividing both sides by $$ar^2\;(\neq0)$$ gives \[r^2=4.\quad(2)\]

2. Solving for the common ratio $$r$$

Equation (2) gives $$r=2\quad\text{or}\quad r=-2.$$

3. Finding the corresponding first term $$a$$

(i) If $$r=2$$, substitute in (1):

$$a(1+2)=-4\;\Longrightarrow\;3a=-4\;\Longrightarrow\;a=-\dfrac{4}{3}.$$

(ii) If $$r=-2$$, substitute in (1):

$$a(1-2)=-4\;\Longrightarrow\;-a=-4\;\Longrightarrow\;a=4.$$

4. Writing the two possible G.P.s

• For $$a=-\dfrac43,\,r=2$$:
  $$-\dfrac43,\,-\dfrac83,\,-\dfrac{16}3,\,-\dfrac{32}3,\,-\dfrac{64}3,\ldots$$
• For $$a=4,\,r=-2$$:
  $$4,\,-8,\,16,\,-32,\,64,\ldots$$

Both progressions satisfy the two given conditions, so they are the required answers.

Step 1 : Set–up an algebraic model

Suppose 100 is written as the sum of n consecutive natural numbers. Let the first (smallest) term be $$a$$. The numbers are then $$a,\;a+1,\;a+2,\;\ldots,\;a+(n-1).$$

Their sum, by the well-known formula for an arithmetic progression, is

Here this sum must equal 100, so from (1)

$$\frac{n}{2}\bigl(2a+n-1\bigr)=100$$

Multiplying both sides by 2 gives the key equation

Step 2 : Convert the problem to a factor–pair search

Equation (2) shows that n must be a positive divisor of 200. For every such divisor, define

$$2a+n-1 = \frac{200}{n}$$ \[\Rightarrow\; a = \frac{200/n\; -\; n\; +\;1}{2}.\quad(3)\]

Because we want consecutive natural numbers, both $$n\ge 1$$ and $$a\ge 1$$ must hold, and the right-hand side of (3) has to be a positive integer.

Step 3 : List all divisors of 200

The positive divisors of 200 are

1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200.

Step 4 : Test each divisor using formula (3)

Step 5 : Write the actual sequences

1. n=1, a=100 → 100
2. n=5, a=18 → 18 + 19 + 20 + 21 + 22 = 100
3. n=8, a=9 → 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100

No other divisor of 200 produces a positive integer value for $$a$$, so the above list is complete.

Conclusion

Hence, 100 can be expressed as a sum of consecutive natural numbers in exactly three different ways (including the trivial one–term case).

Given: At the start (time 0 h) the culture contains 30 bacteria.
Every hour the population doubles.

When a quantity doubles every equal interval of time, it is multiplied by 2 each interval. Hence, after every completed hour the number of bacteria is obtained by multiplying the previous count by 2.

Step 1 – Express the growth law algebraically

If $$N_0$$ is the initial count and $$N_t$$ the count after $$t$$ hours, then

$$N_t = N_0 \times 2^{t}$$   (because the factor 2 is applied $$t$$ times).

Here $$N_0 = 30$$, so

$$N_t = 30 \times 2^{t}.$$

Step 2 – Find the required counts

1. At the end of the 2^nd hour (that is, $$t = 2$$):
   $$N_2 = 30 \times 2^{2} = 30 \times 4 = 120.$$
   Therefore, 120 bacteria are present.

2. At the end of the 4^th hour (that is, $$t = 4$$):
   $$N_4 = 30 \times 2^{4} = 30 \times 16 = 480.$$
   Therefore, 480 bacteria are present.

3. At the end of the $$n^{\text{th}}$$ hour (general case):
   \[N_n = 30 \times 2^{n}.\] Thus, after $$n$$ hours the culture contains $$30 \times 2^{n}$$ bacteria.

Hence, the population follows the geometric progression
30, 60, 120, 240, 480, … corresponding to $$t = 0, 1, 2, 3, 4, \dots$$ hours respectively.

Let the first term ( a ) and common difference ( d ) of the AP be unknown.

The general term is $$T_n = a + (n-1)d$$.

Forming the two given equations

• 4^th term: $$T_4 = a + 3d$$
  8^th term: $$T_8 = a + 7d$$
  Given that their sum is 24:
  $$T_4 + T_8 = 24 \[0.2cm] (a + 3d) + (a + 7d) = 24 \[0.2cm] 2a + 10d = 24$$
  Dividing by 2 gives the first linear relation:
  $$a + 5d = 12 \quad(1)$$
• 6^th term: $$T_6 = a + 5d$$
  10^th term: $$T_{10} = a + 9d$$
  Their sum is 44:
  $$T_6 + T_{10} = 44 \[0.2cm] (a + 5d) + (a + 9d) = 44 \[0.2cm] 2a + 14d = 44$$
  Dividing by 2 gives the second linear relation:
  $$a + 7d = 22 \quad(2)$$

Solving the simultaneous equations

Subtract (1) from (2):

$$\bigl(a + 7d\bigr) - \bigl(a + 5d\bigr) = 22 - 12$$

$$2d = 10 \;\;\Rightarrow\;\; d = 5$$

Substitute $$d = 5$$ in (1):

$$a + 5(5) = 12 \;\;\Rightarrow\;\; a + 25 = 12 \;\;\Rightarrow\;\; a = -13$$

First three terms

The terms are obtained with $$T_n = a + (n-1)d$$:

• 1^st term: $$T_1 = a = -13$$
• 2^nd term: $$T_2 = a + d = -13 + 5 = -8$$
• 3^rd term: $$T_3 = a + 2d = -13 + 10 = -3$$

Therefore, the first three terms of the AP are

Let $$S_n$$ denote the sum of the first $$n$$ natural numbers.

We know the standard formula:

$$S_n = 1 + 2 + 3 + \dots + n = \frac{n\,(n+1)}{2}$$

We want this sum to be greater than 1 000, so we write the inequality

$$\frac{n(n+1)}{2} > 1000$$

Multiply both sides by 2 to clear the denominator:

$$n(n+1) > 2000$$

Expand the left side to get a quadratic inequality:

$$n^2 + n - 2000 > 0$$

First, find the roots of the corresponding quadratic equation $$n^2 + n - 2000 = 0$$ using the quadratic formula.

The discriminant is

$$\Delta = b^2 - 4ac = 1^2 - 4(1)(-2000) = 1 + 8000 = 8001$$

Hence the positive root is

$$n = \frac{-1 + \sqrt{8001}}{2}$$

Compute an approximate value of $$\sqrt{8001}$$:

$$89^2 = 7921 \quad \text{and} \quad 90^2 = 8100$$
Thus $$\sqrt{8001} \approx 89.45$$ (to two decimal places).

So

$$n \approx \frac{-1 + 89.45}{2} \approx \frac{88.45}{2} \approx 44.225$$

The quadratic opens upwards, so the expression $$n^2 + n - 2000$$ is positive for all $$n$$ greater than this root. Therefore, the smallest integer $$n$$ that satisfies the inequality must be the next whole number after 44.225, namely $$n = 45$$.

Verification

• For $$n = 44$$: $$S_{44} = \frac{44\times45}{2} = 990 \lt 1000$$
• For $$n = 45$$: $$S_{45} = \frac{45\times46}{2} = 1035 \gt 1000$$

Thus the smallest value of $$n$$ that makes the sum exceed 1 000 is 45.

Step 1 – Identify the first term and the common ratio

The given geometric progression (GP) is

$$2,\;8,\;32,\;\ldots$$

First term:
$$a = 2$$

Common ratio:
$$r = \frac{8}{2} = 4$$

Step 2 – Write the explicit (general) formula for the $$n^{\text{th}}$$ term

For any GP, if the first term is $$a$$ and the common ratio is $$r$$, the $$n^{\text{th}}$$ term $$a_n$$ is

$$a_n = a\,r^{\,n-1}$$

Substituting $$a = 2$$ and $$r = 4$$:

Step 3 – Find the value of $$n$$ for which $$a_n = 131072$$

Set $$a_n$$ equal to $$131072$$ and use formula (1):

$$2\,\times 4^{\,n-1} = 131072$$

Divide by $$2$$ on both sides:

$$4^{\,n-1} = \frac{131072}{2} = 65536$$

Express $$65536$$ as a power of $$4$$. Listing a few powers:

• $$4^5 = 1024$$
• $$4^6 = 4096$$
• $$4^7 = 16384$$
• $$4^8 = 65536$$

Thus $$4^{\,8} = 65536$$, so

$$n - 1 = 8\quad\Longrightarrow\quad n = 9$$

Hence $$131072$$ is the 9^th term of the GP.

Step 4 – Summarise the formulas

• Explicit formula: $$a_n = 2\,\times 4^{\,n-1}$$
• Recursive formula:
  • Initial term: $$a_1 = 2$$
  • Recursion rule (for $$n \ge 2$$): $$a_n = 4\,a_{n-1}$$

Let the first three terms of the G.P. be $$a,\;ar,\;ar^{2}$$ where $$a$$ is the first term and $$r$$ is the common ratio.

Step 1 : Translate the statements into equations

Sum of the first three terms:

$$a+ar+ar^{2}=\frac{13}{12}$$

Product of the first three terms:

$$a\,\cdot\,ar\,\cdot\,ar^{2}=a^{3}r^{3}=-1$$

Step 2 : Simplify the product equation

From $$a^{3}r^{3}=-1$$ we get

$$(ar)^{3}=-1\;\;\Longrightarrow\;\;ar=-1$$

Hence

$$a=\frac{-1}{r}$$

Step 3 : Substitute $$a=\dfrac{-1}{r}$$ in the sum equation

$$\frac{-1}{r}\bigl(1+r+r^{2}\bigr)=\frac{13}{12}$$

Multiply both sides by $$r$$:

$$-(1+r+r^{2})=\frac{13r}{12}$$

Clear the denominator by multiplying by $$12$$:

$$-12(1+r+r^{2})=13r$$

Remove the brackets:

$$-12-12r-12r^{2}=13r$$

Bring every term to one side:

$$-12r^{2}-25r-12=0$$

Multiply by $$-1$$ to obtain a standard quadratic:

$$12r^{2}+25r+12=0$$

Step 4 : Solve the quadratic for $$r$$

The discriminant is

$$\Delta=25^{2}-4\cdot12\cdot12=625-576=49$$

Hence

$$r=\frac{-25\pm7}{2\cdot12}=\frac{-25\pm7}{24}$$

Therefore

$$r_{1}=\frac{-25+7}{24}=\frac{-18}{24}=-\frac{3}{4},\qquad r_{2}=\frac{-25-7}{24}=\frac{-32}{24}=-\frac{4}{3}$$

Step 5 : Find the corresponding first terms and the three terms

• For $$r=-\dfrac{3}{4}$$:
    $$a=\frac{-1}{r}=\frac{-1}{-3/4}=\frac{4}{3}$$
    Three terms: $$\frac{4}{3},\;\frac{4}{3}\Bigl(-\frac{3}{4}\Bigr)=-1,\;-1\Bigl(-\frac{3}{4}\Bigr)=\frac{3}{4}$$
• For $$r=-\dfrac{4}{3}$$:
    $$a=\frac{-1}{r}=\frac{-1}{-4/3}=\frac{3}{4}$$
    Three terms: $$\frac{3}{4},\;\frac{3}{4}\Bigl(-\frac{4}{3}\Bigr)=-1,\;-1\Bigl(-\frac{4}{3}\Bigr)=\frac{4}{3}$$

Conclusion

The common ratio can be $$-\dfrac{3}{4}$$ or $$-\dfrac{4}{3}$$. Accordingly the first three terms of the G.P. are

• $$\dfrac{4}{3},\;-1,\;\dfrac{3}{4}$$  when  $$r=-\dfrac{3}{4}$$, or
• $$\dfrac{3}{4},\;-1,\;\dfrac{4}{3}$$  when  $$r=-\dfrac{4}{3}$$.

Given: In a geometric progression (GP),

• the 4^th term is $$x$$,
• the 10^th term is $$y$$,
• the 16^th term is $$z$$.

Let the first term of the GP be $$a$$ and the common ratio be $$r$$.

Step 1: Express each given term in algebraic form

The n^th term $$T_n$$ of a GP is $$T_n = a r^{n-1}$$.

Therefore:

• 4^th term: $$T_4 = a r^{4-1} = a r^{3} = x$$     …(1)
• 10^th term: $$T_{10} = a r^{10-1} = a r^{9} = y$$   …(2)
• 16^th term: $$T_{16} = a r^{16-1} = a r^{15} = z$$ …(3)

Step 2: Show that $$y^2 = xz$$

Square equation (2):

$$y^2 = (a r^{9})^2 = a^2 r^{18}$$ …(4)

Multiply equations (1) and (3):

$$xz = (a r^{3})(a r^{15}) = a^2 r^{18}$$ …(5)

From (4) and (5): $$y^2 = a^2 r^{18} = xz$$.

Step 3: Conclude that $$x, y, z$$ form a GP

For any three positive numbers to be in GP, the middle term squared must equal the product of the other two, i.e. $$y^2 = xz$$.

Since we have proved precisely this relationship, the numbers $$x, y, z$$ are in geometric progression.

Hence proved.

Let the first three terms of the required geometric progression (GP) be

$$a, \; ar, \; ar^2$$ where $$a$$ is the first term and $$r$$ is the common ratio.

Step 1 : Use the information about their sum

$$a + ar + ar^2 = 26 \;\Rightarrow\; a(1 + r + r^2) = 26 \quad(1)$$

Step 2 : Use the information about the sum of their squares

$$a^2 + a^2 r^2 + a^2 r^4 = 364 \;\Rightarrow\; a^2(1 + r^2 + r^4) = 364 \quad(2)$$

Step 3 : Eliminate $$a$$

From (1) we have $$a = \dfrac{26}{1 + r + r^2}$$. Substituting this in (2):

$$\left(\dfrac{26}{1 + r + r^2}\right)^2 \bigl(1 + r^2 + r^4\bigr) = 364$$

\[ \frac{676\,(1 + r^2 + r^4)}{(1 + r + r^2)^2} = 364 \]

Dividing by $$13^2=169$$ gives

$$\frac{1 + r^2 + r^4}{(1 + r + r^2)^2} = \frac{91}{169}$$

Cross–multiplying,

$$169(1 + r^2 + r^4) = 91(1 + r + r^2)^2$$

Expand the right-hand side:

$$(1 + r + r^2)^2 = 1 + r^2 + r^4 + 2r + 2r^2 + 2r^3$$

Hence

$$169(1 + r^2 + r^4) = 91\bigl(1 + r^2 + r^4 + 2r + 2r^2 + 2r^3\bigr)$$

Collecting like terms and bringing everything to the left:

$$3r^4 - 7r^3 - 4r^2 - 7r + 3 = 0 \quad(3)$$

Step 4 : Solve the quartic

Equation (3) is palindromic (coefficients read the same forwards and backwards), so divide by $$r^2$$ and set $$x = r + \dfrac{1}{r}$$:

$$3\bigl(r^2 + \frac{1}{r^2}\bigr) - 7\bigl(r + \frac{1}{r}\bigr) - 4 = 0$$

Since $$r^2 + \dfrac{1}{r^2} = x^2 - 2$$ we get

$$3(x^2 - 2) - 7x - 4 = 0 \;\Rightarrow\; 3x^2 - 7x - 10 = 0$$

The discriminant is $$(-7)^2 - 4\times 3\times(-10) = 49 + 120 = 169 = 13^2$$, so

$$x = \frac{7 \pm 13}{6} \;\Rightarrow\; x_1 = \frac{10}{3}, \; x_2 = -1$$

Case I: $$r + \dfrac{1}{r} = \frac{10}{3}$$

Multiplying by $$r$$, $$3r^2 - 10r + 3 = 0$$

The discriminant is $$64$$, giving $$r = \dfrac{10 \pm 8}{6} = 3 \text{ or } \dfrac13$$.

Case II: $$r + \dfrac{1}{r} = -1$$ gives $$r^2 + r + 1 = 0$$ which has no real roots (its discriminant is negative). Hence we discard this case for real GPs.

Step 5 : Find $$a$$ for each admissible $$r$$

• For $$r = 3$$ : $$1 + r + r^2 = 1 + 3 + 9 = 13$$, so $$a = \dfrac{26}{13} = 2$$.
• For $$r = \dfrac13$$ : $$1 + r + r^2 = 1 + \dfrac13 + \dfrac19 = \dfrac{13}{9}$$, so $$a = \dfrac{26}{13/9} = 18$$.

Step 6 : Write the two possible GPs

• If $$r = 3$$: $$2,\; 2\times 3 = 6,\; 6\times 3 = 18$$.
• If $$r = \dfrac13$$: $$18,\; 18\times \dfrac13 = 6,\; 6\times \dfrac13 = 2$$.

In both cases the conditions of the problem are satisfied (check: $$2^2 + 6^2 + 18^2 = 4 + 36 + 324 = 364$$).

Hence the three terms of the GP are either $$2,\;6,\;18$$ or $$18,\;6,\;2$$.

Step 1 : Using the given rule to obtain the first few terms

The rule is

$$P_1 = 1, \; P_2 = 2,$$

and for every integer $$n > 2$$

$$P_n = P_1 + P_2 + \cdots + P_{n-1} + 1.$$

Compute term by term.

• For $$n = 3$$ :
    $$P_3 = P_1 + P_2 + 1 = 1 + 2 + 1 = 4.$$
• For $$n = 4$$ :
    $$P_4 = P_1 + P_2 + P_3 + 1 = 1 + 2 + 4 + 1 = 8.$$
• For $$n = 5$$ :
    $$P_5 = P_1 + P_2 + P_3 + P_4 + 1 = 1 + 2 + 4 + 8 + 1 = 16.$$
• For $$n = 6$$ :
    $$P_6 = 1 + 2 + 4 + 8 + 16 + 1 = 32.$$
• For $$n = 7$$ :
    $$P_7 = 1 + 2 + 4 + 8 + 16 + 32 + 1 = 64.$$
• For $$n = 8$$ :
    $$P_8 = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 1 = 128.$$

Thus

$$P_1 = 1,\; P_2 = 2,\; P_3 = 4,\; P_4 = 8,\; P_5 = 16,\; P_6 = 32,\; P_7 = 64,\; P_8 = 128.$$

Step 2 : Finding a simpler recursive relation

Notice that for $$n > 2$$

\[ P_n - P_{n-1} = \bigl(P_1 + \cdots + P_{n-1} + 1\bigr)\; -\; \bigl(P_1 + \cdots + P_{n-2} + 1\bigr) = P_{n-1}. \]

Therefore

$$P_n = 2P_{n-1}\quad (n \ge 2).$$

So the whole sequence can be generated just by doubling the previous term.

Step 3 : Deriving an explicit formula

Because $$P_n = 2P_{n-1}$$ with the first term $$P_1 = 1,$$ the sequence is geometric with common ratio $$2$$. Hence

\[ P_n = 1 \times 2^{\,n-1} = 2^{n-1}\quad \text{for every positive integer } n. \]

Given :  $$W_1 = 1$$,  $$W_2 = 2$$ and for every integer $$n > 2$$

$$W_n = W_1 + W_2 + \ldots + W_{n-2} + 2.$$

We have to evaluate $$W_1,\,W_2,\ldots ,W_8$$ and then identify the sequence.

Step 1 – Write down the first two terms.

• $$W_1 = 1$$
• $$W_2 = 2$$

Step 2 – Find the next terms using the given rule.

1. For $$n = 3$$
   $$W_3 = W_1 + 2 = 1 + 2 = 3$$
2. For $$n = 4$$
   $$W_4 = W_1 + W_2 + 2 = 1 + 2 + 2 = 5$$
3. For $$n = 5$$
   $$W_5 = W_1 + W_2 + W_3 + 2 = 1 + 2 + 3 + 2 = 8$$
4. For $$n = 6$$
   $$W_6 = W_1 + W_2 + W_3 + W_4 + 2 = 1 + 2 + 3 + 5 + 2 = 13$$
5. For $$n = 7$$
   $$W_7 = W_1 + W_2 + W_3 + W_4 + W_5 + 2 = 1 + 2 + 3 + 5 + 8 + 2 = 21$$
6. For $$n = 8$$
   $$W_8 = W_1 + W_2 + W_3 + W_4 + W_5 + W_6 + 2 = 1 + 2 + 3 + 5 + 8 + 13 + 2 = 34$$

Step 3 – Tabulate the results.

Step 4 – Spotting the pattern.

Observe that from $$n \ge 3$$ each term equals the sum of the two immediately preceding terms:

$$\begin{aligned}W_3 &= 1 + 2,\\ W_4 &= 2 + 3,\\ W_5 &= 3 + 5,\\ &\;\vdots\end{aligned}$$

This is exactly the rule that generates the famous Fibonacci sequence. Our list starts 1, 2, 3, 5, 8, 13, 21, 34, which is the Fibonacci sequence except that the second “1” has been replaced by “2”.

Hence $$\{W_n\}$$ is a Fibonacci‐type sequence.