Chapter 5: Arithmetic Progressions

The given sequence is $$\frac{3}{2},\; \frac{1}{2},\; -\frac{1}{2},\; -\frac{3}{2}, \ldots$$

In an arithmetic progression (AP):

• The first term is denoted by $$a$$.
• The common difference is denoted by $$d$$ and is found by subtracting any term from the term that follows it.

Step 1 – First term

The first entry of the sequence is $$\frac{3}{2}$$, so

$$a = \frac{3}{2}$$

Step 2 – Common difference

Subtract the first term from the second term:

$$d = \frac{1}{2} - \frac{3}{2} = -\frac{2}{2} = -1$$

Verification (optional but reassuring):

$$-\frac{1}{2} - \frac{1}{2} = -1, \qquad -\frac{3}{2} - \bigl(-\frac{1}{2}\bigr) = -1$$

The difference is consistently $$-1$$, confirming the value of $$d$$.

Conclusion

The arithmetic progression has first term $$a = \frac{3}{2}$$ and common difference $$d = -1$$.

The sequence is $$2, 7, 12, \ldots$$.

Compare any two consecutive terms to verify the common difference:

$$d = 7 - 2 = 5$$

Since the difference is constant, the sequence is an arithmetic progression (A.P.).

In an A.P., the n-th term $$a_n$$ is given by

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

Here:

• First term: $$a_1 = 2$$
• Common difference: $$d = 5$$

To find the 10th term, substitute $$n = 10$$:

$$a_{10} = 2 + (10 - 1) \times 5$$

Compute the product and sum step by step:

$$a_{10} = 2 + 9 \times 5$$

$$a_{10} = 2 + 45$$

$$a_{10} = 47$$

Therefore, the 10th term of the given A.P. is $$47$$.

Step 1 : Identify the first term and common difference
Given AP : $$21, 18, 15, \ldots$$
First term $$a = 21$$
Common difference $$d = 18 - 21 = -3$$

Step 2 : Write the general (nth) term
For any AP, $$a_n = a + (n-1)d$$.
Therefore here
$$a_n = 21 + (n-1)(-3) = 21 - 3(n-1).$$

Part A : Find which term is $$-81$$
Set $$a_n = -81$$ and solve for $$n$$:
$$21 - 3(n-1) = -81$$
Subtract 21 from both sides:
$$-3(n-1) = -102$$
Divide by $$-3$$:
$$n-1 = 34$$
Add 1:
$$n = 35$$
So, the 35^th term of the AP is $$-81$$.

Part B : Is any term equal to $$0$$?
Put $$a_n = 0$$:
$$21 - 3(n-1) = 0$$
Subtract 21:
$$-3(n-1) = -21$$
Divide by $$-3$$:
$$n-1 = 7$$
Add 1:
$$n = 8$$
Because $$n = 8$$ is a positive integer, the 8^th term exists and is $$0$$.

Conclusion
The 35^th term is $$-81$$, and yes—there is a term equal to $$0$$, namely the 8^th term.

Let the first term of the required A.P. be $$a$$ and the common difference be $$d$$.

The general (nth) term of an arithmetic progression is

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

1. Information about the third term

For $$n = 3$$, we have

$$a_3 = a + (3-1)d = a + 2d$$

Given that $$a_3 = 5$$, therefore

$$a + 2d = 5 \qquad (1)$$

2. Information about the seventh term

For $$n = 7$$, we have

$$a_7 = a + (7-1)d = a + 6d$$

Given that $$a_7 = 9$$, therefore

$$a + 6d = 9 \qquad (2)$$

3. Solve the simultaneous equations

Subtract equation (1) from equation (2):

$$ (a + 6d) - (a + 2d) = 9 - 5 $$

$$ 4d = 4 $$

$$ d = 1 $$

Substitute $$d = 1$$ back into equation (1):

$$ a + 2(1) = 5 $$

$$ a + 2 = 5 $$

$$ a = 3 $$

4. Write the required A.P.

The first term is $$a = 3$$ and the common difference is $$d = 1$$, so the arithmetic progression is

$$3,\; 4,\; 5,\; 6,\; 7,\; 8,\; 9,\; \ldots$$

Verification:

• Third term: $$a + 2d = 3 + 2(1) = 5$$ ✓
• Seventh term: $$a + 6d = 3 + 6(1) = 9$$ ✓

Hence, the required arithmetic progression is $$3,\;4,\;5,\;6,\;\ldots$$.

Step 1 · Identify the parameters of the progression
For the list $$5,\;11,\;17,\;23,\;\ldots$$ the first term is
$$a = 5$$
and the common difference is
$$d = 11 - 5 = 6$$

Step 2 · Write the general term
The $$n^{\text{th}}$$ term $$a_n$$ of an arithmetic progression is
$$a_n = a + (n-1)d$$

Step 3 · Set the general term equal to 301
To check whether 301 is a term, solve
$$a + (n-1)d = 301$$
Substitute $$a = 5$$ and $$d = 6$$:
$$5 + (n-1)\,6 = 301$$

Step 4 · Solve for n
$$5 + 6(n-1) = 301$$
$$6(n-1) = 301 - 5$$
$$6(n-1) = 296$$
$$n - 1 = \frac{296}{6}$$
$$n - 1 = \frac{148}{3} = 49\tfrac{1}{3}$$
$$n = 50\tfrac{1}{3}$$

Step 5 · Interpret the result
The value of $$n$$ obtained is not an integer. Since the term number must be a positive integer, 301 cannot occur in the progression.

Conclusion
301 is not a term of the arithmetic progression $$5,\;11,\;17,\;23,\;\ldots$$

Let us determine all two-digit numbers that are divisible by 3.

The smallest two-digit number is 10. To reach the first multiple of 3, divide 10 by 3:

$$10 \div 3 = 3\text{ remainder }1$$

Since the remainder is 1, add $$3-1 = 2$$ to 10:

$$10 + 2 = 12$$

Therefore the first (smallest) two-digit multiple of 3 is $$a = 12$$.

The greatest two-digit number is 99, and $$99 \div 3 = 33$$ exactly, so $$l = 99$$ is also divisible by 3.

Thus the required numbers form the arithmetic progression

$$12,\,15,\,18,\,\dots,\,99$$

For an A.P. the nth term is $$a_n = a + (n-1)d$$, where $$a = 12$$, the common difference $$d = 3$$ and the last term $$a_n = 99$$.

Substituting:

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

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

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

$$n-1 = \frac{87}{3} = 29$$

$$n = 29 + 1 = 30$$

Hence, there are $$n = 30$$ two-digit numbers that are divisible by 3.

The terms form an arithmetic progression (AP):

$$a_1 = 10, \; d = 7 - 10 = -3$$

The given last term is $$a_n = -62$$.

1. Find the total number of terms.
   

   For an AP, $$a_n = a_1 + (n-1)d$$. Substitute the known values:

   $$-62 = 10 + (n-1)(-3)$$

   $$-62 = 10 - 3(n-1)$$

   $$-62 = 10 - 3n + 3$$

   $$-62 = 13 - 3n$$

   $$-62 - 13 = -3n$$

   $$-75 = -3n$$

   $$n = \frac{-75}{-3} = 25$$

   So, the AP contains $$25$$ terms.

2. Locate the required term.
   

   The last term itself is counted as the 1st term from the end. Hence, the 11th term from the end is

   $$\bigl(25 - (11 - 1)\bigr)^{\text{th}} = (25 - 10)^{\text{th}} = 15^{\text{th}}$$ term from the beginning.

3. Find the 15th term.
   

   $$a_{15} = a_1 + (15 - 1)d$$

   $$a_{15} = 10 + 14(-3)$$

   $$a_{15} = 10 - 42 = -32$$

Therefore, the 11th term from the last term (towards the first term) is $$-32$$.

Principal invested: $$P = \mathrm{\rlap{/}{R}}\,1000$$
Rate of simple interest: $$R = 8\%\text{ per year}$$

For simple interest, the total interest after $$n$$ years is

$$I_n = \dfrac{P R n}{100}$$

Substituting the given data,

$$I_n = \dfrac{1000 \times 8 \times n}{100} = 80 n$$

Thus, at the end of successive years we obtain

• 1st year: $$I_1 = 80 \times 1 = \mathrm{\rlap{/}{R}}\,80$$
• 2nd year: $$I_2 = 80 \times 2 = \mathrm{\rlap{/}{R}}\,160$$
• 3rd year: $$I_3 = 80 \times 3 = \mathrm{\rlap{/}{R}}\,240$$
• and so on.

The sequence of interests is therefore

$$80,\;160,\;240,\;\dots$$

Common difference:

$$d = 160 - 80 = 80$$

Because the common difference is constant, these interests form an arithmetic progression (AP) with
first term $$a = 80$$ and common difference $$d = 80$$.

Interest at the end of 30 years is the 30-th term $$a_{30}$$ of this AP:

$$a_{30} = a + (30 - 1)d = 80 + 29 \times 80 = 80 + 2320 = \mathrm{\rlap{/}{R}}\,2400$$

Hence, the simple interest accumulated at the end of 30 years is $$\mathrm{\rlap{/}{R}}\,2400$$.

Given the numbers of rose plants in successive rows are

$$23, 21, 19, \ldots , 5$$

Each term decreases by $$2$$, so the sequence is an arithmetic progression (A.P.) with

first term $$a = 23$$ and common difference $$d = -2$$.

Let $$n$$ be the required number of rows. For an A.P., the nth (last) term is

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

Here $$l = 5$$, hence

$$5 = 23 + (n-1)(-2)$$

$$5 = 23 - 2(n-1)$$

$$2(n-1) = 18$$

$$n - 1 = 9$$

$$n = 10$$

Therefore, the flower bed has $$10$$ rows of rose plants.

An arithmetic progression (AP) has nth term given by $$a_n = a + (n-1)d$$ where $$a$$ is the first term and $$d$$ the common difference.

(i) $$a = 7,\; d = 3,\; n = 8$$
$$a_8 = 7 + (8-1)\,3 = 7 + 7\times 3 = 28$$

(ii) $$a = -18,\; n = 10,\; a_{10} = 0$$
$$0 = -18 + (10-1)d \Rightarrow 0 = -18 + 9d \Rightarrow 9d = 18 \Rightarrow d = 2$$

(iii) $$d = -3,\; n = 18,\; a_{18} = -5$$
$$-5 = a + (18-1)(-3) = a - 51 \Rightarrow a = -5 + 51 = 46$$

(iv) $$a = -18.9,\; d = 2.5,\; a_n = 3.6$$
$$3.6 = -18.9 + (n-1)2.5 \Rightarrow (n-1)2.5 = 22.5 \Rightarrow n-1 = \frac{22.5}{2.5} = 9 \Rightarrow n = 10$$

(v) $$a = 3.5,\; d = 0,\; n = 105$$
$$a_{105} = 3.5 + (105-1)\times 0 = 3.5$$

Completed table:

The given arithmetic progression (AP) is $$3,\;8,\;13,\;18,\ldots$$

First term: $$a = 3$$

Common difference: $$d = 8 - 3 = 5$$

Let the term equal to $$78$$ be the $$n^{\text{th}}$$ term, so $$a_n = 78$$.

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

Substituting the known values:

$$78 = 3 + (n - 1)5$$

Subtract 3 from both sides:

$$78 - 3 = 5(n - 1)$$   ⇒   $$75 = 5(n - 1)$$

Divide by 5:

$$15 = n - 1$$

Add 1 to both sides:

$$n = 16$$

Hence, $$78$$ is the $$16^{\text{th}}$$ term of the given AP.

Consider the given arithmetic progression (AP)

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

Step 1 – Identify a and d

• First term: $$a = 11$$
• Common difference: $$d = 8 - 11 = -3$$

Step 2 – Write the general term

The $$n^{\mathrm{th}}$$ term of an AP is $$a_n = a + (n-1)d$$.

Step 3 – Set $$a_n = -150$$ and solve for $$n$$

$$11 + (n-1)(-3) = -150$$

\( (n-1)(-3) = -150 - 11 \)

\( (n-1)(-3) = -161 \)

Divide by $$-3$$:

$$n - 1 = \frac{-161}{-3} = \frac{161}{3}$$

Add 1 to both sides:

$$n = 1 + \frac{161}{3} = \frac{164}{3}$$

Step 4 – Interpret the result

$$\frac{164}{3}$$ is not an integer, so there is no positive integer $$n$$ satisfying the equation.

Conclusion

Because no positive integer value of $$n$$ exists, $$-150$$ is not a term of the given arithmetic progression.

Let the first term and the common difference of the required A.P. be $$a$$ and $$d$$ respectively.

The general (nth) term of an A.P. is given by $$a_n = a + (n-1)d$$.

Use the given data

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

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

Find the common difference $$d$$

Subtract (1) from (2): $$\bigl(a + 15d\bigr) - \bigl(a + 10d\bigr) = 73 - 38$$

This gives $$5d = 35 \;\Rightarrow\; d = 7$$

Find the first term $$a$$

Put $$d = 7$$ in (1): $$a + 10 \times 7 = 38 \;\Rightarrow\; a = 38 - 70 = -32$$

Obtain the 31st term

31st term: $$a_{31} = a + 30d$$

Therefore $$a_{31} = -32 + 30 \times 7 = -32 + 210 = 178$$

Hence, the 31st term of the A.P. is $$178$$.

The nth term of an arithmetic progression (AP) is given by

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

where $$a$$ is the first term and $$d$$ is the common difference.

Step 1 : Translate the given information into equations.

• 3rd term is 12  →  $$a_3 = a + 2d = 12$$ … (1)
• 50th (last) term is 106  →  $$a_{50} = a + 49d = 106$$ … (2)

Step 2 : Find the common difference $$d$$.

Subtract (1) from (2):

$$(a + 49d) - (a + 2d) = 106 - 12$$

$$47d = 94$$

$$d = \frac{94}{47} = 2$$

Step 3 : Determine the first term $$a$$.

Insert $$d = 2$$ in equation (1):

$$a + 2\times 2 = 12 \;\Rightarrow\; a + 4 = 12$$

$$a = 12 - 4 = 8$$

Step 4 : Compute the 29th term.

For $$n = 29$$,

$$a_{29} = a + (29 - 1)d = 8 + 28\times 2 = 8 + 56 = 64$$

Hence, the 29th term of the AP is 64.

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

Using the general term of an AP, the $$n^{\text{th}}$$ term is given by
$$t_n = a + (n-1)d$$.

Information given

• 3^rd term: $$t_3 = a + 2d = 4$$ … (1)
• 9^th term: $$t_9 = a + 8d = -8$$ … (2)

Step 1: Find the common difference $$d$$

Subtract (1) from (2):

$$ (a + 8d) - (a + 2d) = -8 - 4 $$

$$ 6d = -12 $$

$$ d = -2 $$

Step 2: Find the first term $$a$$

Substitute $$d = -2$$ in (1):

$$ a + 2(-2) = 4 $$

$$ a - 4 = 4 $$

$$ a = 8 $$

Step 3: Locate the term that equals zero

Let the required term be the $$n^{\text{th}}$$ term. Set $$t_n = 0$$:

$$ a + (n-1)d = 0 $$

Substitute $$a = 8$$ and $$d = -2$$:

$$ 8 + (n-1)(-2) = 0 $$

$$ 8 - 2(n-1) = 0 $$

$$ 8 - 2n + 2 = 0 $$

$$ 10 - 2n = 0 $$

$$ 2n = 10 $$

$$ n = 5 $$

Conclusion

The 5^th term of the given arithmetic progression is zero.

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

The general or $$n^{\text{th}}$$ term of an AP is given by

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

Step 1 | Write the 17^th and 10^th terms

• 17^th term: $$a_{17} = a + (17-1)d = a + 16d$$
• 10^th term: $$a_{10} = a + (10-1)d = a + 9d$$

Step 2 | Translate the given condition into an equation

“The 17^th term exceeds the 10^th term by 7” means

$$a_{17} - a_{10} = 7$$

Substitute the expressions from Step 1:

$$(a + 16d) - (a + 9d) = 7$$

Step 3 | Simplify

First remove the brackets:

$$a + 16d - a - 9d = 7$$

Cancel the $$a$$ terms:

$$16d - 9d = 7$$

$$7d = 7$$

Step 4 | Solve for d

$$d = \frac{7}{7} = 1$$

Hence, the common difference of the AP is $$1$$.

First recognise the data of the arithmetic progression (AP).

• First term: $$a = 3$$
• Common difference: $$d = 15-3 = 27-15 = 39-27 = 12$$

The general or n-th term of an AP is

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

1. Find the 54th term.

Substitute $$n = 54$$:

$$a_{54} = 3 + (54-1)\,12 = 3 + 53\times12 = 3 + 636 = 639$$

2. Form the required term.

The term we are looking for is “132 more than the 54th term”. Therefore

$$\text{required term value} = 639 + 132 = 771$$

3. Let this required term be the $$n$$-th term; write its formula.

$$a_n = 3 + (n-1)\,12 = 771$$

4. Solve for $$n$$.

$$\begin{aligned} 3 + (n-1)\,12 &= 771 \\ (n-1)\,12 &= 771 - 3 = 768 \\ n-1 &= \frac{768}{12} = 64 \\ n &= 64 + 1 = 65 \end{aligned}$$

5. Conclusion. The 65^th term of the AP is 132 more than its 54^th term.

Let the first terms of the two arithmetic progressions (APs) be $$a$$ and $$b$$, and let their common difference be $$d$$ (the same for both).

The $$n^{\text{th}}$$ term of an AP is given by

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

100^th terms

• First AP: $$T_{100}^{(1)} = a + 99d,$$
• Second AP: $$T_{100}^{(2)} = b + 99d.$$

According to the question, the difference of these 100^th terms is 100:

$$T_{100}^{(1)} - T_{100}^{(2)} = 100$$
$$\Rightarrow (a + 99d) - (b + 99d) = 100$$
$$\Rightarrow a - b = 100.$$

1000^th terms

• First AP: $$T_{1000}^{(1)} = a + 999d,$$
• Second AP: $$T_{1000}^{(2)} = b + 999d.$$

Their difference is

$$T_{1000}^{(1)} - T_{1000}^{(2)} = (a + 999d) - (b + 999d) = a - b.$$

But we already found $$a - b = 100$$. Hence,

$$T_{1000}^{(1)} - T_{1000}^{(2)} = 100.$$

Therefore, the difference between their 1000^th terms is 100.

The three-digit numbers begin at 100 and end at 999.

The required numbers form an arithmetic progression (A.P.) whose

• first term $$a$$ is the smallest three-digit multiple of 7,
• common difference $$d$$ is $$7$$ (successive multiples differ by 7),
• last term $$l$$ is the greatest three-digit multiple of 7.

1. First term
Divide 100 by 7:

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

The next multiple is obtained by adding the difference $$7$$ once more:
$$7\times15 = 105$$.
Hence $$a = 105$$.

2. Last term
Divide 999 by 7:

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

The preceding multiple is obtained by subtracting the remainder from 999:
$$999-5 = 994 = 7\times142$$.
Hence $$l = 994$$.

3. Number of terms
For an A.P., the nth term formula is

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

Set $$a_n = l$$ and solve for $$n$$:

$$105 + (n-1)7 = 994$$
$$(n-1)7 = 994-105 = 889$$
$$n-1 = 889 \div 7 = 127$$
$$n = 127 + 1 = 128$$

Therefore, there are 128 three-digit numbers that are divisible by 7.

Step 1 : Identify the first and last relevant multiples of 4
The multiples of 4 are $$4, 8, 12, 16, \dots$$
• The first multiple of 4 that is greater than 10 is $$12$$.
• The last multiple of 4 that is less than 250 is $$248$$ (because $$250\div4 = 62.5$$ is not an integer).
Hence the required numbers form the arithmetic progression (AP)
$$12, 16, 20, \dots, 248$$
with first term $$a = 12$$ and common difference $$d = 4$$.

Step 2 : Use the nth-term formula
For an AP, $$a_n = a + (n-1)d$$.
Taking $$a_n = 248,\; a = 12,\; d = 4$$:
$$248 = 12 + (n-1)\times4$$
$$248 - 12 = (n-1)\times4$$
$$236 = (n-1)\times4$$
$$n - 1 = \frac{236}{4} = 59$$
$$n = 59 + 1 = 60$$.

Step 3 : Conclusion
Therefore, there are 60 multiples of 4 lying between 10 and 250.

Let the two arithmetic progressions be

AP 1: $$63,\;65,\;67,\;\ldots$$

AP 2: $$3,\;10,\;17,\;\ldots$$

Step 1: Identify the first term and common difference of each AP

For AP 1: first term $$A = 63$$, common difference $$d_1 = 65 - 63 = 2$$.

For AP 2: first term $$B = 3$$, common difference $$d_2 = 10 - 3 = 7$$.

Step 2: Write the general (n^th) term

In any AP, $$T_n = a + (n-1)d$$.

AP 1: $$T_n^{(1)} = 63 + (n-1)\times 2$$

AP 2: $$T_n^{(2)} = 3 + (n-1)\times 7$$

Step 3: Set the two n^th terms equal

$$63 + (n-1)\times 2 = 3 + (n-1)\times 7$$

Step 4: Solve for $$n$$

Simplify each side:

Left: $$63 + 2(n-1) = 63 + 2n - 2 = 61 + 2n$$

Right: $$3 + 7(n-1) = 3 + 7n - 7 = 7n - 4$$

Equate the two:

$$61 + 2n = 7n - 4$$

Subtract $$2n$$ from both sides:

$$61 = 5n - 4$$

Add $$4$$ to both sides:

$$65 = 5n$$

Divide by $$5$$:

$$n = 13$$

Conclusion

The n^th terms of the two APs are equal when $$n = 13$$. Hence, the 13^th term of each progression is the same.

Step 1 · Express the given information

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

• Third term: $$T_3 = a + 2d = 16$$   …(1)
• Seventh term: $$T_7 = a + 6d$$
  Fifth term: $$T_5 = a + 4d$$
  Given: $$T_7 - T_5 = 12$$

Step 2 · Use the condition on the 7th and 5th terms

$$(a + 6d) - (a + 4d) = 12$$

$$2d = 12$$

$$d = 6$$

Step 3 · Find the first term

Substitute $$d = 6$$ in equation (1):

$$a + 2(6) = 16$$

$$a + 12 = 16$$

$$a = 4$$

Step 4 · Write the required AP

The AP with first term $$4$$ and common difference $$6$$ is
$$4, 10, 16, 22, 28, 34, 40, \dots$$

General term: $$a_n = 4 + (n - 1) \times 6$$.

The given list $$3, 8, 13, \ldots , 253$$ is an arithmetic progression (AP).

Step 1 Identify the first term and the common difference.

• First term: $$a = 3$$
• Common difference: $$d = 8 - 3 = 5$$

Step 2 Confirm the last term and the total number of terms (optional check).

The general term of an AP is $$a_n = a + (n-1)d$$. Set $$a_n = 253$$ and solve for $$n$$:

$$253 = 3 + (n-1)\,5$$

$$253 - 3 = 5(n-1)$$   ⇒   $$250 = 5(n-1)$$

$$n-1 = 50$$   ⇒   $$n = 51$$

So the AP has $$51$$ terms and $$253$$ is indeed the 51st (last) term.

Step 3 Interpret “20th term from the last”.

Counting backward, the last term itself is the 1st from the end, the previous term is the 2nd from the end, and so on. Therefore, the 20th term from the end is the $$51-20+1 = 32$$nd term when counted from the beginning.

Step 4 Find the 32nd term from the start.

Use $$a_{32} = a + (32-1)d$$:

$$a_{32} = 3 + 31\times5$$

$$a_{32} = 3 + 155 = 158$$

Alternative one-line method (treating the backward AP):
Backward AP has first term $$l = 253$$ and common difference $$-5$$. The nth term from the end is
$$l_n = l - (n-1)d = 253 - (20-1)\times5 = 253 - 95 = 158$$, confirming the same result.

Hence, the 20th term from the last term of the AP is $$158$$.

Let the first term of the arithmetic progression be $$a$$ and the common difference be $$d$$.
Then the general term is $$a_n=a+(n-1)d$$.

Form the two given conditions

• 4th term: $$a_4=a+3d$$
• 8th term: $$a_8=a+7d$$

Their sum is given to be 24:

$$a_4+a_8=(a+3d)+(a+7d)=2a+10d=24 \quad\Rightarrow\quad a+5d=12\quad(1)$$

• 6th term: $$a_6=a+5d$$
• 10th term: $$a_{10}=a+9d$$

Their sum is given to be 44:

$$a_6+a_{10}=(a+5d)+(a+9d)=2a+14d=44 \quad\Rightarrow\quad a+7d=22\quad(2)$$

Solve the simultaneous equations

Subtract (1) from (2):

$$\bigl(a+7d\bigr)-(a+5d)=22-12 \;\Rightarrow\; 2d=10 \;\Rightarrow\; d=5.$$

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

$$a+5\times5=12 \;\Rightarrow\; a+25=12 \;\Rightarrow\; a=-13.$$

First three terms

$$a_1=a=-13,\qquad a_2=a+d=-13+5=-8,\qquad a_3=a+2d=-13+10=-3.$$

Let the salary in each successive year form a sequence:

Year 1 (1995): $$a = 5000$$
Year 2 (1996): $$5000 + 200 = 5200$$
Year 3 (1997): $$5200 + 200 = 5400$$, and so on.

Because the amount added every year is the same, the salaries constitute an arithmetic progression (AP) with

• first term $$a = 5000$$
• common difference $$d = 200$$

Let $$n$$ be the year number (counting 1995 as $$n = 1$$) in which the salary is $$7000$$. For an AP, the general (nth) term is

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

Substituting the known values:

$$7000 = 5000 + (n-1)\times 200$$

Subtract $$5000$$ from both sides:

$$2000 = (n-1)\times 200$$

Divide by $$200$$:

$$n - 1 = \frac{2000}{200} = 10$$

Therefore, $$n = 11$$.

Thus, the salary becomes $$7000$$ in the 11^th year of service. Year 1 corresponds to 1995, so

Year 11 = 1995 + 10 = 2005.

The income reached $$\mathrm{\rlap{/}{R}}\,7000$$ in the year 2005.

Let Ramkali’s weekly savings form the sequence

$$5,\;5+1.75,\;5+2\times1.75,\;\ldots$$

Because a fixed amount is added every week, the sequence is an Arithmetic Progression (A.P.).

First term:

$$a = 5$$

Common difference:

$$d = 1.75$$

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

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

We are told that the n-th week’s saving is $$20.75$$, so

$$a_n = 20.75$$

Substitute $$a = 5$$, $$d = 1.75$$ and $$a_n = 20.75$$ into the formula:

$$20.75 = 5 + (n-1)\times1.75$$

Isolate the bracket term:

$$20.75 - 5 = (n-1)\times1.75$$

$$15.75 = (n-1)\times1.75$$

Divide both sides by $$1.75$$ to solve for $$(n-1)$$:

$$n-1 = \frac{15.75}{1.75}$$

Calculate the quotient:

$$\frac{15.75}{1.75} = 9$$  (because $$1.75\times9 = 15.75$$)

Therefore,

$$n - 1 = 9 \;\Longrightarrow\; n = 9 + 1 = 10$$

Hence, Ramkali’s weekly saving becomes Rs 20.75 in the 10^th week.

The given numbers $$8, 3, -2,\ldots$$ form an arithmetic progression (AP).

For any AP let

• first term be $$a_1$$,
• common difference be $$d$$,
• number of terms be $$n$$,
• sum of the first $$n$$ terms be $$S_n$$.

From the question:

First term: $$a_1 = 8$$

Common difference: $$d = 3 - 8 = -5$$

Number of terms: $$n = 22$$

The formula for the sum of the first $$n$$ terms is

$$S_n = \frac{n}{2}\,[2a_1 + (n-1)d]$$

Substitute the known values:

$$S_{22} = \frac{22}{2}\,[2 \times 8 + (22-1)(-5)]$$

Compute each part step by step:

$$\frac{22}{2} = 11$$

$$2 \times 8 = 16$$

$$(22 - 1) = 21 \implies 21 \times (-5) = -105$$

$$16 + (-105) = -89$$

Therefore, $$S_{22} = 11 \times (-89) = -979$$

Thus, the sum of the first 22 terms of the AP is $$-979$$.

Let the first term be denoted by $$a$$ and the common difference by $$d$$.

Given $$a = 10$$ and the sum of the first 14 terms $$S_{14}=1050$$.

The sum of the first $$n$$ terms of an arithmetic progression is

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

Substituting $$n = 14$$, $$a = 10$$ into the formula:

$$1050 = \dfrac{14}{2}\,[2(10) + (14-1)d]$$

$$1050 = 7\,[20 + 13d]$$

Divide by 7:

$$20 + 13d = 150$$

$$13d = 150 - 20 = 130$$

$$d = \dfrac{130}{13} = 10$$

Now, the 20^th term is

$$a_{20} = a + (20-1)d$$

$$a_{20} = 10 + 19 \times 10 = 10 + 190 = 200$$

Thus, the 20^th term of the given arithmetic progression is $$200$$.

We are given the arithmetic progression (AP)

24, 21, 18, \ldots

Its first term and common difference are

$$a = 24, \; d = 21-24 = -3$$

Let $$n$$ be the number of terms whose sum is required to be 78. For an AP the sum of the first $$n$$ terms is

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

Setting $$S_n = 78$$ gives

$$\frac{n}{2}\,[2(24) + (n-1)(-3)] = 78$$

Simplify the bracket first:

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

$$2(24) + (-3n + 3) = 48 + 3 - 3n = 51 - 3n$$

Substitute back:

$$\frac{n}{2}\,(51 - 3n) = 78$$

Multiply by 2:

$$n(51 - 3n) = 156$$

Expand and rearrange to get a quadratic equation:

$$51n - 3n^2 = 156$$

$$-3n^2 + 51n - 156 = 0$$

Multiply by $$-1$$ and divide by 3:

$$n^2 - 17n + 52 = 0$$

Factorisation:

$$n^2 - 13n - 4n + 52 = 0$$

$$(n-13)(n-4) = 0$$

Hence

$$n = 13 \;\text{or}\; n = 4$$

Both values are positive integers, so each is admissible. (Indeed, \(S_4 = 24+21+18+15 = 78\) and \(S_{13} = 78\) as well.)

Therefore, the required number of terms can be either 4 or 13.

We are given the general or $$n$$th term

$$a_n = 3 + 2n$$

To write the progression, substitute successive natural numbers for $$n$$:

• For $$n = 1$$, $$a_1 = 3 + 2(1) = 5$$
• For $$n = 2$$, $$a_2 = 3 + 2(2) = 7$$
• For $$n = 3$$, $$a_3 = 3 + 2(3) = 9$$

Thus the list begins $$5,\,7,\,9,\dots$$. The common difference is

$$d = a_2 - a_1 = 7 - 5 = 2$$

Hence it is an arithmetic progression (AP) with

• first term $$a = 5$$
• common difference $$d = 2$$
• number of terms required $$n = 24$$

The sum of the first $$n$$ terms of an AP is

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

Substitute $$a = 5$$, $$d = 2$$ and $$n = 24$$:

$$ S_{24} = \frac{24}{2}\,[2 \times 5 + (24-1)\times 2] $$

Simplify step by step:

$$\frac{24}{2} = 12$$

$$2 \times 5 = 10$$

$$(24-1)\times 2 = 23 \times 2 = 46$$

So

$$S_{24} = 12\,(10 + 46) = 12 \times 56 = 672$$

Therefore, the sum of the first 24 terms of the given sequence is

Let the required number of terms be $$n$$.
For the given AP we have first term $$a = 9$$ and common difference $$d = 8$$.

The sum of the first $$n$$ terms of an AP is

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

According to the question, $$S_n = 636$$, therefore

$$\dfrac{n}{2}\,[2 \times 9 + (n-1)\times 8] = 636$$

Simplifying the bracket:

$$18 + 8n - 8 = 8n + 10$$

So

$$\dfrac{n}{2}(8n + 10) = 636$$

Because $$8n + 10 = 2(4n + 5)$$, this can be written as

$$n(4n + 5) = 636$$

$$4n^{2} + 5n - 636 = 0$$

Compute the discriminant:

$$\Delta = 5^{2} + 4\times 4 \times 636 = 25 + 10176 = 10201 = 101^{2}$$

Applying the quadratic formula:

$$n = \dfrac{-5 \pm 101}{2\times 4}$$

Ignoring the negative value,

$$n = \dfrac{96}{8} = 12$$

Therefore, 12 terms of the AP must be taken so that their sum is 636.

We are told that the first term, the last term and the sum of an arithmetic progression (AP) are

• First term: $$a = 5$$
• Last term: $$l = 45$$
• Sum of all terms: $$S_n = 400$$

Let $$n$$ be the number of terms and $$d$$ the common difference. We must determine both $$n$$ and $$d$$.

1. Finding the number of terms $$n$$

The sum of the first $$n$$ terms of an AP whose first term is $$a$$ and whose last term is $$l$$ is

$$S_n = \frac{n}{2}\,(a + l).$$

Substituting the given values:

$$400 = \frac{n}{2}\,(5 + 45).$$

Simplify the bracket:

$$400 = \frac{n}{2}\,(50).$$

Multiply both sides by $$2$$ to clear the denominator:

$$800 = n\,(50).$$

Divide by $$50$$:

$$n = \frac{800}{50} = 16.$$

Hence the AP contains $$16$$ terms.

2. Finding the common difference $$d$$

The nth (last) term of an AP is also given by

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

Substitute $$l = 45$$, $$a = 5$$ and $$n = 16$$:

$$45 = 5 + (16 - 1)d.$$

Compute the bracket:

$$45 = 5 + 15d.$$

Subtract $$5$$ from both sides:

$$40 = 15d.$$

Finally divide by $$15$$:

$$d = \frac{40}{15} = \frac{8}{3}.$$

Conclusion

The arithmetic progression has

• Number of terms: $$n = 16$$
• Common difference: $$d = \dfrac{8}{3}$$.

The first term is $$a = 17$$, the common difference is $$d = 9$$ and the last term is $$l = 350$$.

The last term is the $$n^{\text{th}}$$ term, so

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

Substituting the known numbers:

$$350 = 17 + (n-1)9$$

$$350 - 17 = 9(n-1)$$

$$333 = 9(n-1)$$

$$n-1 = \dfrac{333}{9} = 37$$

$$n = 37 + 1 = 38$$

Hence the A.P. contains $$38$$ terms.

The sum of the first $$n$$ terms is $$S_n = \dfrac{n}{2}(a + l)$$.

$$S_{38} = \dfrac{38}{2}(17 + 350) = 19 \times 367 = 6973$$

Therefore, the progression has 38 terms and their sum is 6973.

We are given:

• Common difference $$d = 7$$
• 22^nd term $$a_{22} = 149$$

Let the first term be $$a$$ and the required sum be $$S_{22}$$.

Step 1 Find the first term.

The general term of an AP is $$a_n = a + (n-1)d$$. For $$n = 22$$:

$$a_{22} = a + (22-1)d$$

Substitute the given values:

$$149 = a + 21 \times 7$$

Compute the product:

$$149 = a + 147$$

Solve for $$a$$:

$$a = 149 - 147 = 2$$

Step 2 Find the sum of the first 22 terms.

The sum of the first $$n$$ terms of an AP is $$S_n = \dfrac{n}{2}\,(a + a_n)$$. Here $$n = 22$$, $$a = 2$$, $$a_{22} = 149$$:

$$S_{22} = \dfrac{22}{2}\,(2 + 149)$$

Simplify the fraction:

$$S_{22} = 11\,(151)$$

Multiply:

$$S_{22} = 1661$$

Therefore, the sum of the first 22 terms is $$1661$$.

Let the first term of the A.P. be $$a$$ and the common difference be $$d$$.

Using the formula for the $n$-th term, $$a_n = a + (n-1)d$$:

Second term ⇒ $$a + d = 14$$ …(1)

Third term ⇒ $$a + 2d = 18$$  …(2)

Subtract (1) from (2):

$$\bigl(a + 2d\bigr) - \bigl(a + d\bigr) = d = 18 - 14 = 4$$

Thus, $$d = 4$$.

Substitute $$d$$ in (1):

$$a + 4 = 14 \;\Longrightarrow\; a = 14 - 4 = 10$$

Now, to find the sum of the first $$51$$ terms, use $$S_n = \dfrac{n}{2}\,[2a + (n-1)d]$$ with $$n = 51,\;a = 10,\;d = 4$$.

Compute the bracketed term:

$$2a + (n-1)d = 2(10) + 50(4) = 20 + 200 = 220$$

Hence,

$$S_{51} = \dfrac{51}{2}\times 220$$

$$\displaystyle \;\; = 51 \times 110 = 5610$$

Therefore, the sum of the first 51 terms is $$5610$$.

The sum of the first $$m$$ terms of an AP with first term $$a$$ and common difference $$d$$ is

$$S_m = \frac{m}{2}\bigl(2a + (m-1)d\bigr).$$

Step 1 : Use the given information.

For $$m = 7$$ we have

$$S_7 = \frac{7}{2}\bigl(2a + 6d\bigr) = 49.$$

Multiply by $$\,2\,\,$$ and divide by $$\,7\,\,$$:

$$2a + 6d = 14 \;\;\Longrightarrow\;\; a + 3d = 7 \quad\text{(i)}$$

For $$m = 17$$ we have

$$S_{17} = \frac{17}{2}\bigl(2a + 16d\bigr) = 289.$$

Multiply by $$\,2\,\,$$ and divide by $$\,17\,\,$$:

$$2a + 16d = 34 \;\;\Longrightarrow\;\; a + 8d = 17 \quad\text{(ii)}$$

Step 2 : Solve for $$a$$ and $$d$$.

Subtract (i) from (ii):

$$ (a + 8d) - (a + 3d) = 17 - 7 \;\;\Longrightarrow\;\; 5d = 10 \;\;\Longrightarrow\;\; d = 2. $$

Substitute $$d = 2$$ in (i):

$$ a + 3(2) = 7 \;\;\Longrightarrow\;\; a = 1. $$

Step 3 : General formula for the sum of $$n$$ terms.

$$\begin{aligned} S_n & = \frac{n}{2}\bigl(2a + (n-1)d\bigr) \\[4pt] & = \frac{n}{2}\bigl[2(1) + (n-1)\,2\bigr] \\[4pt] & = \frac{n}{2}\bigl(2 + 2n - 2\bigr) \\[4pt] & = \frac{n}{2}\,(2n) \\[4pt] & = n^2. \end{aligned}$$

Thus, for every natural number $$n$$,

$$S_n = n^2.$$

The sum of the first $$n$$ terms of the AP is given as

$$S_n = 4n - n^2$$  for every natural number $$n$$.

(i) First term (sum of first one term)

Put $$n = 1$$ in the expression for $$S_n$$:

$$S_1 = 4(1) - (1)^2 = 4 - 1 = 3$$

Thus the first term is $$a_1 = 3$$.

(ii) Sum of first two terms

Put $$n = 2$$:

$$S_2 = 4(2) - (2)^2 = 8 - 4 = 4$$

Therefore the sum of the first two terms is $$S_2 = 4$$.

(iii) Second term

The second term is the difference of the first two partial sums:

$$a_2 = S_2 - S_1 = 4 - 3 = 1$$

(iv) Third term

First find $$S_3$$:

$$S_3 = 4(3) - (3)^2 = 12 - 9 = 3$$

Then

$$a_3 = S_3 - S_2 = 3 - 4 = -1$$

(v) Tenth term

Find $$S_{10}$$ and $$S_9$$:

$$S_{10} = 4(10) - (10)^2 = 40 - 100 = -60$$

$$S_9 = 4(9) - (9)^2 = 36 - 81 = -45$$

Hence

$$a_{10} = S_{10} - S_9 = (-60) - (-45) = -15$$

(vi) General (nth) term

In any AP, $$a_n = S_n - S_{n-1}$$. We already have $$S_n$$; now find $$S_{n-1}$$:

$$S_{n-1} = 4(n-1) - (n-1)^2$$

Simplify $$S_{n-1}$$ completely:

$$\begin{aligned} S_{n-1} &= 4n - 4 - \bigl(n^2 - 2n + 1\bigr) \\[2pt] &= 4n - 4 - n^2 + 2n - 1 \\[2pt] &= (4n + 2n) - n^2 - 5 \\[2pt] &= 6n - n^2 - 5. \end{aligned}$$

Now subtract:

$$\begin{aligned} a_n &= S_n - S_{n-1} \\[2pt] &= (4n - n^2) - \bigl(6n - n^2 - 5\bigr) \\[2pt] &= 4n - n^2 - 6n + n^2 + 5 \\[2pt] &= -2n + 5. \end{aligned}$$

Therefore the general term (nth term) of the AP is

$$a_n = 5 - 2n.$$

Quick check: $$a_1 = 5 - 2(1) = 3$$, $$a_2 = 5 - 2(2) = 1$$, $$a_3 = 5 - 2(3) = -1$$, agreeing with the results obtained above. The common difference is $$d = a_2 - a_1 = -2$$, which is consistent throughout.

The positive integers divisible by 6 form the sequence

$$6,\;12,\;18,\;24,\;\ldots$$

This is an arithmetic progression (AP) because each term exceeds the previous one by a constant difference.

Identify the AP parameters

• First term: $$a = 6$$
• Common difference: $$d = 6$$
• Number of required terms: $$n = 40$$ ("first 40" integers)

Formula for the sum of the first $$n$$ terms of an AP

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

Substitute $$a = 6,\;d = 6,\;n = 40$$:

$$S_{40} = \frac{40}{2}\,[\,2(6) + (40-1)(6)\,]$$

Simplify step by step:

1. Calculate $$\frac{40}{2}$$: $$\frac{40}{2} = 20$$.
2. Compute $$2(6)$$: $$2(6) = 12$$.
3. Find $$40-1$$: $$40-1 = 39$$.
4. Multiply $$39 \times 6$$: $$39 \times 6 = 234$$.
5. Add: $$12 + 234 = 246$$.

Now multiply:

$$S_{40} = 20 \times 246 = 4920$$

Therefore, the sum of the first 40 positive integers divisible by 6 is $$4920$$.

The first 15 multiples of 8 are:

$$8,\,16,\,24,\,32,\;\ldots\;,\,120$$

This list is an arithmetic progression (A.P.) because each term increases by the constant common difference $$d = 8$$.

Identify the parameters of the A.P.:

• First term $$a = 8$$
• Common difference $$d = 8$$
• Number of terms $$n = 15$$

For an arithmetic progression the sum of the first $$n$$ terms is

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

Substitute the known values:

$$S_{15} = \frac{15}{2}\,[\,2(8) + (15-1)(8)\,]$$

Compute step by step:

$$2(8) = 16$$

$$(15-1) = 14$$

$$14\times8 = 112$$

$$2a + (n-1)d = 16 + 112 = 128$$

$$S_{15} = \frac{15}{2}\times128$$

$$128 \times 15 = 1920$$

$$\frac{1920}{2} = 960$$

Hence, the sum of the first 15 multiples of 8 is $$960$$.

All odd natural numbers between 0 and 50 form the list

$$1,\;3,\;5,\;\ldots,\;49$$

This list is an arithmetic progression (AP) because the difference between consecutive terms is constant.

First term: $$a = 1$$

Common difference: $$d = 3 - 1 = 2$$

Last term: $$l = 49$$

To find the number of terms $$n$$, use the nth-term formula of an AP:

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

Substituting $$l = 49,\;a = 1,\;d = 2$$,

$$49 = 1 + (n-1)\times 2$$

$$49 - 1 = 2(n-1)$$

$$48 = 2(n-1)$$

$$\frac{48}{2} = n - 1$$

$$24 = n - 1$$

$$n = 25$$

Now use the sum formula for the first $$n$$ terms of an AP:

$$S_n = \frac{n}{2}\,(a + l)$$

Substituting $$n = 25,\;a = 1,\;l = 49$$,

$$S_{25} = \frac{25}{2}\,(1 + 49)$$

$$S_{25} = \frac{25}{2}\times 50$$

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

$$S_{25} = 625$$

Therefore, the sum of all odd numbers between 0 and 50 is $$625$$.

Given penalties per day form the sequence
$$200,\;250,\;300,\;\dots$$

The common difference is constant:

$$d = 250 - 200 = 50, \; 300 - 250 = 50 \;\Rightarrow\; d = 50$$

Thus the penalties constitute an arithmetic progression (AP) with
first term $$a = 200$$ and common difference $$d = 50$$.

The contractor delayed the work by $$n = 30$$ days. We need the sum $$S_{30}$$ of the first 30 terms of the AP.

Step 1   Find the 30^th (last) term

The nth term of an AP is $$a_n = a + (n-1)d$$. For $$n = 30$$,

$$l = a_{30} = 200 + (30-1)\times 50 = 200 + 29\times 50 = 200 + 1450 = 1650.$$

Step 2   Sum of the first 30 terms

The sum of the first $$n$$ terms of an AP is $$S_n = \dfrac{n}{2}(a + l)$$. Substitute $$n = 30, \; a = 200, \; l = 1650$$:

$$S_{30} = \dfrac{30}{2}(200 + 1650) = 15 \times 1850 = 27\,750.$$

Therefore, the total penalty the contractor has to pay is
$$\mathrm{\rlap{/}{R}}\,27\,750$$.

Let the value (in rupees) of the first cash prize be $$a$$. Because each successive prize is $$\mathrm{\rlap{/}{R}}\,20$$ less than the preceding one, the seven prizes form an arithmetic progression (A.P.)

$$a,\;a-20,\;a-40,\;a-60,\;a-80,\;a-100,\;a-120$$

For an A.P.:

• first term $$a_1=a$$
• common difference $$d=-20$$
• number of terms $$n=7$$
• sum of terms $$S_n=700$$

The sum of the first $$n$$ terms of an A.P. is given by

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

Substituting the known values:

$$700=\frac{7}{2}\,[2a+6(-20)]$$

$$700=\frac{7}{2}\,(2a-120)$$

Multiply both sides by 2:

$$1400=7\,(2a-120)$$

Divide by 7:

$$200=2a-120$$

Add 120 to both sides:

$$320=2a$$

Hence

$$a=160$$

Therefore, the seven prizes are

$$160,\;140,\;120,\;100,\;80,\;60,\;40$$

(all amounts are in rupees).

The number of trees a single section of Class I, II, …, XII will plant is the same as the class number:

Class I → 1 tree    Class II → 2 trees    …    Class XII → 12 trees

Since every class has three sections, the total trees planted by one entire class are thrice the class number.

Thus the numbers of trees for the 12 classes are

$$3,\;6,\;9,\;12,\;\ldots,\;36$$

This list is an arithmetic progression (A.P.) with

• first term $$a = 3$$
• common difference $$d = 6-3 = 3$$
• number of terms $$n = 12$$ (Classes I to XII)
• last term $$l = 36$$

Sum of the first $$n$$ terms of an A.P. is $$S_n = \dfrac{n}{2}\,(a + l).$$

Therefore

$$S_{12} = \frac{12}{2}\,(3 + 36) = 6 \times 39 = 234.$$

Hence, the students will plant a total of 234 trees.

Step 1 : Identify the radii of the 13 semicircles
The radii increase by 0.5 cm each time:

$$r_1 = 0.5\text{ cm}, \; r_2 = 1.0\text{ cm}, \; r_3 = 1.5\text{ cm}, \; \ldots$$
Thus $$r_n = 0.5n\text{ cm},\; n = 1,2,3,\ldots,13.$$

Step 2 : Write the length of each semicircle
Length of a semicircle of radius $$r$$ is half the circumference of a full circle:

$$l_n = \pi r_n = \pi(0.5n) = 0.5\pi n\text{ cm}.$$

Step 3 : Form the series for the total length
For 13 semicircles,

$$L = l_1+l_2+\cdots+l_{13} = 0.5\pi(1+2+\cdots+13).$$

Step 4 : Evaluate the sum of the first 13 natural numbers
$$1+2+\cdots+13 = \frac{13\times14}{2}=91.$$

Step 5 : Substitute and simplify
$$\begin{aligned}L &= 0.5\pi\times91\\ &= 45.5\pi\text{ cm}.\end{aligned}$$

Step 6 : Put $$\pi=\frac{22}{7}$$
$$\begin{aligned}L &= 45.5\times\frac{22}{7}\\ &= 45.5\times3.142857\ldots\\ &= 143\text{ cm}.\end{aligned}$$

Therefore, the total length of the spiral formed by 13 semicircles is $$143\text{ cm}$$.

Let the number of rows be $$n$$.

The numbers of logs in successive rows are

$$20,\;19,\;18,\;\ldots$$

This is an arithmetic progression (AP) with

• first term $$a = 20$$,
• common difference $$d = -1$$.

The total number of logs is the sum of the first $$n$$ terms of the AP and is given to be 200.

Sum of $$n$$ terms:

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

Substituting $$S_n = 200,\;a = 20,\;d = -1$$:

$$200 = \frac{n}{2}\,[2\times 20 + (n-1)(-1)]$$

Simplify inside the bracket:

$$2\times 20 = 40, \quad (n-1)(-1) = -n + 1$$

Hence

$$200 = \frac{n}{2}\,[40 - n + 1] = \frac{n}{2}\,(41 - n).$$

Multiply by 2:

$$400 = n(41 - n).$$

Rearrange to get a quadratic equation:

$$-n^{2} + 41n - 400 = 0 \;\;\Longrightarrow\;\; n^{2} - 41n + 400 = 0.$$

Factorise:

$$n^{2} - 41n + 400 = (n - 16)(n - 25) = 0.$$

Therefore $$n = 16$$ or $$n = 25$$.

If $$n = 25$$, the number of logs in the top row would be

$$a + (n-1)d = 20 + 24(-1) = -4,$$

which is impossible. Hence $$n = 16$$.

Logs in the top row:

$$a + (n-1)d = 20 + 15(-1) = 5.$$

Therefore, the logs are arranged in 16 rows and the top row contains 5 logs.

The positions of the ten potatoes, measured from the bucket along the straight line, form an arithmetic progression.

• Distance of the 1^st potato from the bucket: $$5\,\text{m}$$
• Potatoes are kept $$3\,\text{m}$$ apart, so each next potato is $$3\,\text{m}$$ farther than the previous one.

Therefore the distances (in metres) of the potatoes from the bucket are

$$5,\; 5+3,\; 5+2\times3,\; \dotsc$$

That is,

$$5,\; 8,\; 11,\; 14,\; 17,\; 20,\; 23,\; 26,\; 29,\; 32$$

This list is an arithmetic progression (A.P.) with

first term $$a = 5,\quad\text{common difference } d = 3,\quad\text{number of terms } n = 10$$

When the competitor picks any one potato she has to run to it and back, i.e. twice its distance from the bucket. Hence, for each potato the distance covered is $$2 \times (\text{its distance from the bucket})$$.

Let $$S_{10}$$ be the sum of the distances of all ten potatoes from the bucket. Using the sum formula for an A.P.,

$$S_{10} = \frac{n}{2}\,[2a + (n-1)d]$$

Substituting $$n = 10,\; a = 5,\; d = 3$$,

$$S_{10} = \frac{10}{2}\,[2\times5 + (10-1)\times3]$$

$$\phantom{S_{10}} = 5\,[10 + 9\times3]$$

$$\phantom{S_{10}} = 5\,[10 + 27] = 5\times37 = 185$$

The total distance run is twice this value, because every potato requires a trip to and fro:

$$\text{Total distance} = 2 \times 185 = 370\,\text{m}$$

Hence, the competitor has to run $$\mathbf{370\,\text{metres}}$$ in all.

The AP is $$121,\;117,\;113,\ldots$$

Identify the first term and common difference:

• First term: $$a = 121$$
• Common difference: $$d = 117 - 121 = -4$$

The general (nth) term of an AP is $$a_n = a + (n-1)d$$.

We want the first term that is negative, i.e. the smallest positive integer $$n$$ for which $$a_n < 0$$.

Substitute $$a$$ and $$d$$ into the expression for $$a_n$$ and set it less than $$0$$:

$$a_n = 121 + (n-1)(-4) < 0$$

Simplify the inequality step by step:

$$121 - 4(n-1) < 0$$

Expand the brackets:

$$121 - 4n + 4 < 0$$

Combine like terms:

$$125 - 4n < 0$$

Rearrange to isolate $$n$$ (remember to reverse the inequality sign if multiplying or dividing by a negative number; here we divide by positive $$4$$, so the sign stays the same):

$$-4n < -125 \;\;\Rightarrow\;\; 4n > 125\;\;\Rightarrow\;\; n > \frac{125}{4}$$

Calculate $$\frac{125}{4} = 31.25$$.

The smallest integer greater than $$31.25$$ is $$32$$. Therefore $$n = 32$$ gives the first negative term.

Verify by actually finding the 32nd term:

$$a_{32} = 121 + (32-1)(-4) = 121 + 31\times(-4) = 121 - 124 = -3$$

Indeed $$a_{32} = -3 < 0$$, while the previous term (the 31st term) would be

$$a_{31} = 121 + 30\times(-4) = 121 - 120 = 1 > 0$$

Hence, the 32nd term is the first negative term of the given AP.

Let $$a$$ be the first term and $$d$$ the common difference of the required arithmetic progression (AP).

The third term is $$T_3 = a + 2d$$ and the seventh term is $$T_7 = a + 6d$$.

Using the given conditions

1. Sum of the two terms:
   $$T_3 + T_7 = 6 \;\Rightarrow\; (a + 2d) + (a + 6d) = 6 \;\Rightarrow\; 2a + 8d = 6 \;\Rightarrow\; a + 4d = 3$$(1)

2. Product of the two terms:
   $$T_3 \times T_7 = 8 \;\Rightarrow\; (a + 2d)(a + 6d) = 8$$(2)

From (1): $$a = 3 - 4d$$.

Substitute in the factors of (2):
$$T_3 = (3 - 4d) + 2d = 3 - 2d,$$
$$T_7 = (3 - 4d) + 6d = 3 + 2d.$$

Hence
$$(3 - 2d)(3 + 2d) = 8 \;\Rightarrow\; 9 - 4d^2 = 8 \;\Rightarrow\; 4d^2 = 1 \;\Rightarrow\; d^2 = \frac{1}{4}.$$

Thus $$d = \frac{1}{2} \text{ or } d = -\frac{1}{2}.$$

Corresponding first terms

• For $$d = \frac{1}{2}$$: $$a = 3 - 4\left(\frac{1}{2}\right) = 1.$$
• For $$d = -\frac{1}{2}$$: $$a = 3 - 4\left(-\frac{1}{2}\right) = 5.$$

Sum of the first 16 terms

The sum of $$n$$ terms of an AP is $$S_n = \frac{n}{2}\,[2a + (n-1)d]$$.
For $$n = 16$$,
$$S_{16} = \frac{16}{2}[2a + 15d] = 8[2a + 15d].$$

Case I: $$a = 1,\; d = \frac{1}{2}$$
$$S_{16} = 8\,[2(1) + 15\left(\frac{1}{2}\right)] = 8\,(2 + 7.5) = 8 \times 9.5 = 76.$$

Case II: $$a = 5,\; d = -\frac{1}{2}$$
$$S_{16} = 8\,[2(5) + 15\left(-\frac{1}{2}\right)] = 8\,(10 - 7.5) = 8 \times 2.5 = 20.$$

Hence, depending on which of the two admissible progressions is chosen, the required sum is either $$76$$ or $$20$$.

The distance between the top and bottom rungs is given in metres but the spacing and the lengths of the rungs are expressed in centimetres, so first change metres to centimetres:

$$2\dfrac{1}{2}\text{ m}=2.5\text{ m}=2.5\times100=250\text{ cm}$$

1. Number of rungs

Every two successive rungs are $$25\text{ cm}$$ apart. The number of equal gaps between the top and bottom rungs is therefore

$$\frac{250}{25}=10$$

Since the first rung produces the first gap, the total number of rungs is one more than the number of gaps:

$$n = 10 + 1 = 11$$

2. Showing that the lengths form an arithmetic progression

The bottom (first) rung is $$45\text{ cm}$$ long and the top (last) rung is $$25\text{ cm}$$ long. The lengths decrease uniformly, so they form an AP whose

• first term $$a = 45\text{ cm}$$,
• last term $$l = 25\text{ cm}$$,
• number of terms $$n = 11$$.

For completeness, the common difference is

$$d = \frac{l-a}{n-1}=\frac{25-45}{11-1}=\frac{-20}{10}=-2\text{ cm}$$,

so the sequence is $$45,\;43,\;41,\ldots,25\;(11\text{ terms}).$$

3. Sum of the lengths

The total length of wood required equals the sum of this AP:

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

Substituting $$n = 11,\;a = 45,\;l = 25$$,

$$S_{11}=\frac{11}{2}(45+25)=\frac{11}{2}(70)=11\times35=385\text{ cm}.$$

4. Converting back to metres (if desired)

$$385\text{ cm}=\frac{385}{100}\text{ m}=3.85\text{ m}.$$

Hence, the carpenter needs $$3\,\text{m}\,85\text{ cm}\;(=3.85\text{ m})$$ of wood for the rungs.

Given : The house numbers form the sequence

$$1,\,2,\,3,\,\ldots ,\,49$$

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

Let the required house number be $$x$$.
Then

• all houses before it are numbered $$1,2,3,\ldots ,x-1$$,
• all houses after it are numbered $$x+1, x+2,\ldots ,49$$.

According to the question,

$$\text{(sum of numbers before }x\text{)} = \text{(sum of numbers after }x\text{)}$$

Using the hint, write this condition with the standard notation $$S_n$$ for the sum of the first $$n$$ terms of an A.P.:

$$S_{x-1} = S_{49} - S_x$$

Step 1 : Formula for $$S_n$$

For any A.P. having first term $$a$$ and common difference $$d,$$ the sum of the first $$n$$ terms is

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

Here $$a = 1,\;d = 1,$$ so this reduces to

$$S_n = \dfrac{n}{2}\bigl[2 \times 1 + (n-1)\times 1\bigr] = \dfrac{n}{2}(n+1) = \dfrac{n(n+1)}{2}$$

Step 2 : Write each sum through the formula

$$S_{x-1} = \dfrac{(x-1)x}{2}$$

$$S_{49} = \dfrac{49\times 50}{2} = 1225$$

$$S_x = \dfrac{x(x+1)}{2}$$

Step 3 : Form the required equation

Substitute these three expressions in $$S_{x-1} = S_{49} - S_x$$:

$$\dfrac{(x-1)x}{2} = 1225 - \dfrac{x(x+1)}{2}$$

Multiply every term by 2 to clear the denominator:

$$x(x-1) = 2450 - x(x+1)$$

Step 4 : Simplify the equation

Move the right-hand product to the left:

$$x(x-1) + x(x+1) = 2450$$

Factor out $$x$$:

$$x\bigl[(x-1) + (x+1)\bigr] = 2450$$

$$x\bigl(2x\bigr) = 2450$$

$$2x^2 = 2450$$

Divide by 2:

$$x^2 = 1225$$

Take the positive square root (house numbers are positive):

$$x = 35$$

Step 5 : Verification

Sum of numbers up to 34:

$$S_{34} = \dfrac{34\times 35}{2} = 595$$

Sum of numbers from 36 to 49:

$$S_{49} - S_{35} = 1225 - \dfrac{35\times 36}{2} = 1225 - 630 = 595$$

The two sums are equal, confirming the result.

Hence, there is a house number for which the required condition holds, and that number is

$$x = 35$$

Given data

• Number of steps = 15
• Length of every step = $$50\text{ m}$$
• Rise (vertical thickness) of each step = $$\dfrac14\text{ m}$$
• Tread (horizontal depth) of each step = $$\dfrac12\text{ m}$$

For one step having height $$\dfrac14\text{ m}$$, depth $$\dfrac12\text{ m}$$ and length $$50\text{ m}$$, the elementary volume is

$$V_0 = \frac14 \times \frac12 \times 50 = 6.25\;\text{m}^3.$$

Why the steps form an A.P.

Observe the shaded vertical slices of width $$\dfrac12\text{ m}$$ (see the hint figure).

• The first slice (front slice) must be filled up to a height of $$1 \times \dfrac14\text{ m}$$.
  Volume of slice 1  = $$1\,V_0 = 6.25\text{ m}^3.$$
• The second slice reaches a height of $$2 \times \dfrac14\text{ m}$$.
  Volume of slice 2  = $$2\,V_0 = 12.5\text{ m}^3.$$
• Similarly, the n-th slice (step) rises to a height of $$n \times \dfrac14\text{ m}$$ and therefore needs
  $$n\,V_0$$ cubic metres of concrete.

Thus the required volumes form the sequence

$$6.25,\;12.5,\;18.75,\;\dots$$

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

$$a = 6.25, \; d = 6.25, \; n = 15.$$

Total volume

The sum of the first $$n$$ terms of an A.P. is
$$S_n = \dfrac{n}{2}\,[\,2a + (n-1)d\,].$$

Substituting the values,

$$\begin{aligned} S_{15} &= \dfrac{15}{2}\,[\,2(6.25) + 14(6.25)\,]\\[4pt] &= \dfrac{15}{2}\,[6.25\times16]\\[4pt] &= \dfrac{15}{2}\times100\\[4pt] &= 15\times50\\[4pt] &= 750. \end{aligned}$$

Therefore, the total volume of concrete needed is $$750\,\text{m}^3.$$