Chapter 8: Sequences and Series

The sequence is defined by $$a_n = (n-1)(2-n)(3+n)$$. To find the 20th term, substitute $$n = 20$$.

$$a_{20} = (20-1)(2-20)(3+20)$$

$$a_{20} = (19)(-18)(23)$$

Multiply step by step. First, $$19 \times (-18) = -342$$. Then $$-342 \times 23 = -7866$$.

Hence the 20th term is $$a_{20} = -7866$$.

We are given $$a_1 = 1$$ and the recurrence $$a_n = a_{n-1} + 2$$ for $$n \geq 2$$. That is, each term is obtained by adding $$2$$ to the previous term.

$$a_1 = 1$$

$$a_2 = a_1 + 2 = 1 + 2 = 3$$

$$a_3 = a_2 + 2 = 3 + 2 = 5$$

$$a_4 = a_3 + 2 = 5 + 2 = 7$$

$$a_5 = a_4 + 2 = 7 + 2 = 9$$

So the first five terms are $$1, 3, 5, 7, 9$$.

The corresponding series is obtained by connecting the terms with plus signs:

$$1 + 3 + 5 + 7 + 9 + \ldots$$

The $$n$$th term is $$a_n = n(n+2)$$. Substitute $$n = 1, 2, 3, 4, 5$$.

$$a_1 = 1(1+2) = 1 \times 3 = 3$$

$$a_2 = 2(2+2) = 2 \times 4 = 8$$

$$a_3 = 3(3+2) = 3 \times 5 = 15$$

$$a_4 = 4(4+2) = 4 \times 6 = 24$$

$$a_5 = 5(5+2) = 5 \times 7 = 35$$

Hence the first five terms are $$3, 8, 15, 24, 35$$.

The $$n$$th term is $$a_n = \dfrac{n}{n+1}$$. Substitute $$n = 1, 2, 3, 4, 5$$.

$$a_1 = \dfrac{1}{1+1} = \dfrac{1}{2}$$

$$a_2 = \dfrac{2}{2+1} = \dfrac{2}{3}$$

$$a_3 = \dfrac{3}{3+1} = \dfrac{3}{4}$$

$$a_4 = \dfrac{4}{4+1} = \dfrac{4}{5}$$

$$a_5 = \dfrac{5}{5+1} = \dfrac{5}{6}$$

Hence the first five terms are $$\dfrac{1}{2}, \dfrac{2}{3}, \dfrac{3}{4}, \dfrac{4}{5}, \dfrac{5}{6}$$.

The $$n$$th term is $$a_n = 2^n$$. Substitute $$n = 1, 2, 3, 4, 5$$.

$$a_1 = 2^1 = 2$$

$$a_2 = 2^2 = 4$$

$$a_3 = 2^3 = 8$$

$$a_4 = 2^4 = 16$$

$$a_5 = 2^5 = 32$$

Hence the first five terms are $$2, 4, 8, 16, 32$$.

The $$n$$th term is $$a_n = \dfrac{2n-3}{6}$$. Substitute $$n = 1, 2, 3, 4, 5$$.

$$a_1 = \dfrac{2(1)-3}{6} = \dfrac{-1}{6} = -\dfrac{1}{6}$$

$$a_2 = \dfrac{2(2)-3}{6} = \dfrac{1}{6}$$

$$a_3 = \dfrac{2(3)-3}{6} = \dfrac{3}{6} = \dfrac{1}{2}$$

$$a_4 = \dfrac{2(4)-3}{6} = \dfrac{5}{6}$$

$$a_5 = \dfrac{2(5)-3}{6} = \dfrac{7}{6}$$

Hence the first five terms are $$-\dfrac{1}{6}, \dfrac{1}{6}, \dfrac{1}{2}, \dfrac{5}{6}, \dfrac{7}{6}$$.

The $$n$$th term is $$a_n = (-1)^{n-1}\, 5^{n+1}$$. The factor $$(-1)^{n-1}$$ makes the terms alternate in sign (it is $$+1$$ for odd $$n$$ and $$-1$$ for even $$n$$). Substitute $$n = 1, 2, 3, 4, 5$$.

$$a_1 = (-1)^{0}\, 5^{2} = 1 \times 25 = 25$$

$$a_2 = (-1)^{1}\, 5^{3} = -1 \times 125 = -125$$

$$a_3 = (-1)^{2}\, 5^{4} = 1 \times 625 = 625$$

$$a_4 = (-1)^{3}\, 5^{5} = -1 \times 3125 = -3125$$

$$a_5 = (-1)^{4}\, 5^{6} = 1 \times 15625 = 15625$$

Hence the first five terms are $$25, -125, 625, -3125, 15625$$.

The $$n$$th term is $$a_n = n\cdot\dfrac{n^2+5}{4}$$. Substitute $$n = 1, 2, 3, 4, 5$$.

$$a_1 = 1\cdot\dfrac{1^2+5}{4} = 1\cdot\dfrac{6}{4} = \dfrac{3}{2}$$

$$a_2 = 2\cdot\dfrac{2^2+5}{4} = 2\cdot\dfrac{9}{4} = \dfrac{18}{4} = \dfrac{9}{2}$$

$$a_3 = 3\cdot\dfrac{3^2+5}{4} = 3\cdot\dfrac{14}{4} = \dfrac{42}{4} = \dfrac{21}{2}$$

$$a_4 = 4\cdot\dfrac{4^2+5}{4} = 4\cdot\dfrac{21}{4} = 21$$

$$a_5 = 5\cdot\dfrac{5^2+5}{4} = 5\cdot\dfrac{30}{4} = \dfrac{150}{4} = \dfrac{75}{2}$$

Hence the first five terms are $$\dfrac{3}{2}, \dfrac{9}{2}, \dfrac{21}{2}, 21, \dfrac{75}{2}$$.

The $$n$$th term is $$a_n = 4n - 3$$.

To find $$a_{17}$$, substitute $$n = 17$$:

$$a_{17} = 4(17) - 3 = 68 - 3 = 65$$

To find $$a_{24}$$, substitute $$n = 24$$:

$$a_{24} = 4(24) - 3 = 96 - 3 = 93$$

Hence $$a_{17} = 65$$ and $$a_{24} = 93$$.

The $$n$$th term is $$a_n = \dfrac{n^2}{2^n}$$.

To find $$a_7$$, substitute $$n = 7$$:

$$a_7 = \dfrac{7^2}{2^7} = \dfrac{49}{128}$$

Hence $$a_7 = \dfrac{49}{128}$$.

The $$n$$th term is $$a_n = (-1)^{n-1}\, n^3$$.

To find $$a_9$$, substitute $$n = 9$$:

$$a_9 = (-1)^{9-1}\, (9)^3 = (-1)^{8}\, (729)$$

Since $$(-1)^{8} = 1$$,

$$a_9 = 1 \times 729 = 729$$

Hence $$a_9 = 729$$.

The $$n$$th term is $$a_n = \dfrac{n(n-2)}{n+3}$$.

To find $$a_{20}$$, substitute $$n = 20$$:

$$a_{20} = \dfrac{20(20-2)}{20+3} = \dfrac{20 \times 18}{23} = \dfrac{360}{23}$$

Hence $$a_{20} = \dfrac{360}{23}$$.

We are given $$a_1 = 3$$ and the recurrence $$a_n = 3a_{n-1} + 2$$ for $$n > 1$$. We compute each term using the one before it.

$$a_1 = 3$$

$$a_2 = 3a_1 + 2 = 3(3) + 2 = 9 + 2 = 11$$

$$a_3 = 3a_2 + 2 = 3(11) + 2 = 33 + 2 = 35$$

$$a_4 = 3a_3 + 2 = 3(35) + 2 = 105 + 2 = 107$$

$$a_5 = 3a_4 + 2 = 3(107) + 2 = 321 + 2 = 323$$

So the first five terms are $$3, 11, 35, 107, 323$$.

The corresponding series is:

$$3 + 11 + 35 + 107 + 323 + \ldots$$

We are given $$a_1 = -1$$ and the recurrence $$a_n = \dfrac{a_{n-1}}{n}$$ for $$n \geq 2$$.

$$a_1 = -1$$

$$a_2 = \dfrac{a_1}{2} = \dfrac{-1}{2} = -\dfrac{1}{2}$$

$$a_3 = \dfrac{a_2}{3} = \dfrac{-1/2}{3} = -\dfrac{1}{6}$$

$$a_4 = \dfrac{a_3}{4} = \dfrac{-1/6}{4} = -\dfrac{1}{24}$$

$$a_5 = \dfrac{a_4}{5} = \dfrac{-1/24}{5} = -\dfrac{1}{120}$$

So the first five terms are $$-1, -\dfrac{1}{2}, -\dfrac{1}{6}, -\dfrac{1}{24}, -\dfrac{1}{120}$$.

The corresponding series is:

$$-1 + \left(-\dfrac{1}{2}\right) + \left(-\dfrac{1}{6}\right) + \left(-\dfrac{1}{24}\right) + \left(-\dfrac{1}{120}\right) + \ldots$$

We are given $$a_1 = a_2 = 2$$ and the recurrence $$a_n = a_{n-1} - 1$$ for $$n > 2$$.

$$a_1 = 2$$

$$a_2 = 2$$

$$a_3 = a_2 - 1 = 2 - 1 = 1$$

$$a_4 = a_3 - 1 = 1 - 1 = 0$$

$$a_5 = a_4 - 1 = 0 - 1 = -1$$

So the first five terms are $$2, 2, 1, 0, -1$$.

The corresponding series is:

$$2 + 2 + 1 + 0 + (-1) + \ldots$$

The Fibonacci sequence has $$a_1 = a_2 = 1$$ and $$a_n = a_{n-1} + a_{n-2}$$ for $$n > 2$$. We first generate enough terms.

$$a_1 = 1,\quad a_2 = 1$$

$$a_3 = a_2 + a_1 = 1 + 1 = 2$$

$$a_4 = a_3 + a_2 = 2 + 1 = 3$$

$$a_5 = a_4 + a_3 = 3 + 2 = 5$$

$$a_6 = a_5 + a_4 = 5 + 3 = 8$$

Now we compute the ratio $$\dfrac{a_{n+1}}{a_n}$$ for each value of $$n$$.

For $$n = 1$$: $$\dfrac{a_2}{a_1} = \dfrac{1}{1} = 1$$

For $$n = 2$$: $$\dfrac{a_3}{a_2} = \dfrac{2}{1} = 2$$

For $$n = 3$$: $$\dfrac{a_4}{a_3} = \dfrac{3}{2}$$

For $$n = 4$$: $$\dfrac{a_5}{a_4} = \dfrac{5}{3}$$

For $$n = 5$$: $$\dfrac{a_6}{a_5} = \dfrac{8}{5}$$

In the given G.P. $$5, 25, 125, \ldots$$ the first term is $$a = 5$$ and the common ratio is

$$r = \dfrac{25}{5} = 5$$

The general term of a G.P. is $$a_n = a r^{n-1}$$.

The 10th term: substitute $$n = 10$$:

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

The $$n$$th term:

$$a_n = a r^{n-1} = 5 \cdot 5^{n-1} = 5^{n}$$

Hence $$a_{10} = 5^{10}$$ and $$a_n = 5^{n}$$.

In the G.P. $$2, 8, 32, \ldots$$ the first term is $$a = 2$$ and the common ratio is

$$r = \dfrac{8}{2} = 4$$

Let the $$n$$th term be $$131072$$. Using $$a_n = a r^{n-1}$$:

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

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

Now express $$65536$$ as a power of $$4$$. Since $$4^8 = 65536$$,

$$4^{n-1} = 4^{8}$$

Equating the exponents,

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

Hence $$131072$$ is the 9th term of the G.P.

Let the first term be $$a$$ and the common ratio be $$r$$. The general term is $$a_n = a r^{n-1}$$.

The 3rd term is $$24$$:

$$a r^{2} = 24 \qquad \text{...(1)}$$

The 6th term is $$192$$:

$$a r^{5} = 192 \qquad \text{...(2)}$$

Dividing equation (2) by equation (1):

$$\dfrac{a r^{5}}{a r^{2}} = \dfrac{192}{24} \quad\Rightarrow\quad r^{3} = 8 \quad\Rightarrow\quad r = 2$$

Substituting $$r = 2$$ in equation (1):

$$a (2)^2 = 24 \quad\Rightarrow\quad 4a = 24 \quad\Rightarrow\quad a = 6$$

The 10th term is

$$a_{10} = a r^{9} = 6 \cdot 2^{9} = 6 \times 512 = 3072$$

In the series $$1 + \dfrac{2}{3} + \dfrac{4}{9} + \ldots$$ the first term is $$a = 1$$ and the common ratio is

$$r = \dfrac{2/3}{1} = \dfrac{2}{3}$$

Since $$r < 1$$, the sum of the first $$n$$ terms is

$$S_n = \dfrac{a(1 - r^n)}{1 - r} = \dfrac{1\left(1 - \left(\frac{2}{3}\right)^n\right)}{1 - \frac{2}{3}} = \dfrac{1 - \left(\frac{2}{3}\right)^n}{\frac{1}{3}}$$

$$S_n = 3\left[1 - \left(\dfrac{2}{3}\right)^n\right]$$

For the sum of the first $$5$$ terms, put $$n = 5$$:

$$S_5 = 3\left[1 - \left(\dfrac{2}{3}\right)^5\right] = 3\left[1 - \dfrac{32}{243}\right] = 3 \cdot \dfrac{211}{243} = \dfrac{211}{81}$$

In the G.P. $$3, \dfrac{3}{2}, \dfrac{3}{4}, \ldots$$ the first term is $$a = 3$$ and the common ratio is

$$r = \dfrac{3/2}{3} = \dfrac{1}{2}$$

Since $$r < 1$$, the sum of $$n$$ terms is

$$S_n = \dfrac{a(1 - r^n)}{1 - r} = \dfrac{3\left(1 - \left(\frac{1}{2}\right)^n\right)}{1 - \frac{1}{2}} = 6\left[1 - \left(\dfrac{1}{2}\right)^n\right]$$

We are told this sum equals $$\dfrac{3069}{512}$$:

$$6\left[1 - \left(\dfrac{1}{2}\right)^n\right] = \dfrac{3069}{512}$$

$$1 - \left(\dfrac{1}{2}\right)^n = \dfrac{3069}{3072}$$

$$\left(\dfrac{1}{2}\right)^n = 1 - \dfrac{3069}{3072} = \dfrac{3}{3072} = \dfrac{1}{1024}$$

Since $$1024 = 2^{10}$$, we have $$\left(\dfrac{1}{2}\right)^n = \left(\dfrac{1}{2}\right)^{10}$$, so

$$n = 10$$

Hence $$10$$ terms are needed.

Let the three terms of the G.P. be $$\dfrac{a}{r}, \ a, \ ar$$. Choosing this symmetric form makes the product simple.

Using the product:

$$\dfrac{a}{r} \cdot a \cdot ar = a^3 = -1 \quad\Rightarrow\quad a = -1$$

Using the sum:

$$\dfrac{a}{r} + a + ar = \dfrac{13}{12}$$

$$a\left(\dfrac{1}{r} + 1 + r\right) = \dfrac{13}{12}$$

Substituting $$a = -1$$:

$$-\left(\dfrac{1}{r} + 1 + r\right) = \dfrac{13}{12} \quad\Rightarrow\quad \dfrac{1}{r} + r = -\dfrac{13}{12} - 1 = -\dfrac{25}{12}$$

Multiplying through by $$12r$$:

$$12 + 12r^2 = -25r \quad\Rightarrow\quad 12r^2 + 25r + 12 = 0$$

Solving this quadratic:

$$r = \dfrac{-25 \pm \sqrt{625 - 576}}{24} = \dfrac{-25 \pm 7}{24}$$

$$r = -\dfrac{18}{24} = -\dfrac{3}{4} \qquad \text{or} \qquad r = -\dfrac{32}{24} = -\dfrac{4}{3}$$

The terms. With $$a = -1$$ and $$r = -\dfrac{3}{4}$$:

$$\dfrac{a}{r} = \dfrac{-1}{-3/4} = \dfrac{4}{3}, \qquad a = -1, \qquad ar = (-1)\left(-\dfrac{3}{4}\right) = \dfrac{3}{4}$$

So the terms are $$\dfrac{4}{3}, -1, \dfrac{3}{4}$$.

With $$a = -1$$ and $$r = -\dfrac{4}{3}$$ the same three numbers occur in reverse order: $$\dfrac{3}{4}, -1, \dfrac{4}{3}$$.

This is not a G.P., but it can be reduced to one. Let

$$S = 7 + 77 + 777 + \ldots \text{ to } n \text{ terms}$$

Take out the common factor $$7$$ and then multiply and divide by $$9$$:

$$S = 7(1 + 11 + 111 + \ldots) = \dfrac{7}{9}(9 + 99 + 999 + \ldots \text{ to } n \text{ terms})$$

Write each term as a power of $$10$$ minus $$1$$, since $$9 = 10 - 1$$, $$99 = 10^2 - 1$$, and so on:

$$S = \dfrac{7}{9}\big[(10 - 1) + (10^2 - 1) + \ldots + (10^n - 1)\big]$$

$$S = \dfrac{7}{9}\big[(10 + 10^2 + \ldots + 10^n) - n\big]$$

The bracketed sum $$10 + 10^2 + \ldots + 10^n$$ is a G.P. with first term $$10$$, common ratio $$10$$ and $$n$$ terms:

$$10 + 10^2 + \ldots + 10^n = \dfrac{10(10^n - 1)}{10 - 1} = \dfrac{10(10^n - 1)}{9}$$

Therefore

$$S = \dfrac{7}{9}\left[\dfrac{10(10^n - 1)}{9} - n\right] = \dfrac{70}{81}(10^n - 1) - \dfrac{7n}{9}$$

The number of ancestors in successive generations is

$$2, \ 4, \ 8, \ \ldots$$

This is a G.P. with first term $$a = 2$$ and common ratio $$r = 2$$ (each generation has twice as many ancestors as the next-closer one).

For ten generations we need the sum of $$n = 10$$ terms:

$$S_{10} = \dfrac{a(r^n - 1)}{r - 1} = \dfrac{2(2^{10} - 1)}{2 - 1}$$

Since $$2^{10} = 1024$$,

$$S_{10} = 2(1024 - 1) = 2 \times 1023 = 2046$$

Hence the person has $$2046$$ ancestors during the ten preceding generations.

Let the three numbers inserted be $$G_1, G_2, G_3$$. Then the sequence

$$1, \ G_1, \ G_2, \ G_3, \ 256$$

is a G.P. with $$5$$ terms. Here the first term is $$a = 1$$ and the 5th term is $$256$$.

Using $$a_5 = a r^{4}$$:

$$1 \cdot r^{4} = 256 \quad\Rightarrow\quad r^{4} = 256$$

Since $$256 = 4^4$$, we get $$r = \pm 4$$.

Case 1: $$r = 4$$.

$$G_1 = ar = 4, \quad G_2 = ar^2 = 16, \quad G_3 = ar^3 = 64$$

Case 2: $$r = -4$$.

$$G_1 = ar = -4, \quad G_2 = ar^2 = 16, \quad G_3 = ar^3 = -64$$

Hence the three numbers are $$4, 16, 64$$ (for $$r = 4$$) or $$-4, 16, -64$$ (for $$r = -4$$).

For two positive numbers $$a$$ and $$b$$, the A.M. is $$\dfrac{a+b}{2}$$ and the G.M. is $$\sqrt{ab}$$.

Given A.M. $$= 10$$:

$$\dfrac{a+b}{2} = 10 \quad\Rightarrow\quad a + b = 20 \qquad \text{...(1)}$$

Given G.M. $$= 8$$:

$$\sqrt{ab} = 8 \quad\Rightarrow\quad ab = 64 \qquad \text{...(2)}$$

We find $$a - b$$ using the identity $$(a-b)^2 = (a+b)^2 - 4ab$$:

$$(a-b)^2 = (20)^2 - 4(64) = 400 - 256 = 144$$

$$a - b = \pm 12$$

Taking $$a - b = 12$$ together with $$a + b = 20$$ and adding:

$$2a = 32 \quad\Rightarrow\quad a = 16, \qquad b = 20 - 16 = 4$$

(Taking $$a - b = -12$$ simply interchanges the numbers.)

Hence the two numbers are $$16$$ and $$4$$.

In the G.P. $$\dfrac{5}{2}, \dfrac{5}{4}, \dfrac{5}{8}, \ldots$$ the first term is $$a = \dfrac{5}{2}$$ and the common ratio is

$$r = \dfrac{5/4}{5/2} = \dfrac{5}{4} \times \dfrac{2}{5} = \dfrac{1}{2}$$

The general term is $$a_n = a r^{n-1}$$.

20th term (put $$n = 20$$):

$$a_{20} = \dfrac{5}{2}\left(\dfrac{1}{2}\right)^{19} = \dfrac{5}{2 \cdot 2^{19}} = \dfrac{5}{2^{20}}$$

$$n$$th term:

$$a_n = \dfrac{5}{2}\left(\dfrac{1}{2}\right)^{n-1} = \dfrac{5}{2 \cdot 2^{n-1}} = \dfrac{5}{2^{n}}$$

Let the first term be $$a$$ and the common ratio be $$r = 2$$. The general term is $$a_n = a r^{n-1}$$.

The 8th term is $$192$$:

$$a_8 = a r^{7} = 192 \qquad \text{...(1)}$$

We need the 12th term:

$$a_{12} = a r^{11}$$

Write $$a r^{11} = (a r^{7}) \cdot r^{4}$$, so using (1):

$$a_{12} = 192 \cdot r^{4} = 192 \cdot 2^{4} = 192 \times 16 = 3072$$

Hence the 12th term is $$3072$$.

Let the first term of the G.P. be $$a$$ and the common ratio be $$r$$. The general term is $$a_n = a r^{n-1}$$.

The 5th, 8th and 11th terms are

$$p = a_5 = a r^{4}, \qquad q = a_8 = a r^{7}, \qquad s = a_{11} = a r^{10}$$

Now compute $$q^2$$:

$$q^2 = (a r^{7})^2 = a^2 r^{14}$$

Next compute $$ps$$:

$$ps = (a r^{4})(a r^{10}) = a^2 r^{4+10} = a^2 r^{14}$$

Since both $$q^2$$ and $$ps$$ equal $$a^2 r^{14}$$, we conclude

$$q^2 = ps$$

Hence proved.

Let the first term be $$a = -3$$ and the common ratio be $$r$$. The general term is $$a_n = a r^{n-1}$$.

The 2nd term is $$a_2 = a r$$ and the 4th term is $$a_4 = a r^{3}$$.

It is given that the 4th term equals the square of the 2nd term:

$$a r^{3} = (a r)^{2} = a^2 r^{2}$$

Dividing both sides by $$a r^{2}$$ (which is non-zero):

$$r = a = -3$$

The 7th term is

$$a_7 = a r^{6} = (-3)(-3)^{6} = (-3)^{7} = -2187$$

Hence the 7th term is $$-2187$$.

Three numbers are in G.P. when the ratio of consecutive terms is the same, that is, the middle term squared equals the product of the outer terms.

For $$-\dfrac{2}{7}, \ x, \ -\dfrac{7}{2}$$ to be in G.P.:

$$\dfrac{x}{-\frac{2}{7}} = \dfrac{-\frac{7}{2}}{x}$$

Cross-multiplying:

$$x^2 = \left(-\dfrac{2}{7}\right)\left(-\dfrac{7}{2}\right) = 1$$

Therefore

$$x = \pm 1$$

Hence the numbers are in G.P. when $$x = 1$$ or $$x = -1$$.

In the G.P. $$0.15, 0.015, 0.0015, \ldots$$ the first term is $$a = 0.15$$ and the common ratio is

$$r = \dfrac{0.015}{0.15} = 0.1$$

Since $$r < 1$$, the sum of $$20$$ terms is

$$S_{20} = \dfrac{a(1 - r^{20})}{1 - r} = \dfrac{0.15\,(1 - (0.1)^{20})}{1 - 0.1}$$

$$S_{20} = \dfrac{0.15}{0.9}\,(1 - (0.1)^{20}) = \dfrac{1}{6}\,(1 - (0.1)^{20})$$

Hence

$$S_{20} = \dfrac{1}{6}\left(1 - \dfrac{1}{10^{20}}\right)$$

In the G.P. $$\sqrt{7}, \sqrt{21}, 3\sqrt{7}, \ldots$$ the first term is $$a = \sqrt{7}$$ and the common ratio is

$$r = \dfrac{\sqrt{21}}{\sqrt{7}} = \sqrt{\dfrac{21}{7}} = \sqrt{3}$$

Since $$r > 1$$, the sum of $$n$$ terms is

$$S_n = \dfrac{a(r^{n} - 1)}{r - 1} = \dfrac{\sqrt{7}\left((\sqrt{3})^{n} - 1\right)}{\sqrt{3} - 1}$$

Rationalise the denominator by multiplying numerator and denominator by $$(\sqrt{3} + 1)$$:

$$S_n = \dfrac{\sqrt{7}\left((\sqrt{3})^{n} - 1\right)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \dfrac{\sqrt{7}\,(\sqrt{3} + 1)\left((\sqrt{3})^{n} - 1\right)}{3 - 1}$$

$$S_n = \dfrac{\sqrt{7}\,(\sqrt{3} + 1)}{2}\left((\sqrt{3})^{n} - 1\right)$$

In the G.P. $$1, -a, a^2, -a^3, \ldots$$ the first term is $$1$$ and the common ratio is

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

The sum of $$n$$ terms of a G.P. is $$S_n = \dfrac{a_1(1 - r^{n})}{1 - r}$$, where $$a_1$$ is the first term. Here $$a_1 = 1$$ and $$r = -a$$:

$$S_n = \dfrac{1\big(1 - (-a)^{n}\big)}{1 - (-a)} = \dfrac{1 - (-a)^{n}}{1 + a}$$

This is valid since $$a \neq -1$$ keeps the denominator $$1 + a$$ non-zero.

In the G.P. $$x^3, x^5, x^7, \ldots$$ the first term is $$a = x^3$$ and the common ratio is

$$r = \dfrac{x^5}{x^3} = x^{2}$$

The sum of $$n$$ terms is

$$S_n = \dfrac{a(r^{n} - 1)}{r - 1} = \dfrac{x^3\big((x^{2})^{n} - 1\big)}{x^{2} - 1}$$

$$S_n = \dfrac{x^3\,(x^{2n} - 1)}{x^{2} - 1}$$

This is valid since $$x \neq \pm 1$$ keeps the denominator $$x^2 - 1$$ non-zero.

Split the summation into two parts:

$$\sum_{k=1}^{11} (2 + 3^k) = \sum_{k=1}^{11} 2 \ + \ \sum_{k=1}^{11} 3^{k}$$

First sum: adding $$2$$ eleven times,

$$\sum_{k=1}^{11} 2 = 2 \times 11 = 22$$

Second sum: $$\sum_{k=1}^{11} 3^{k} = 3 + 3^2 + 3^3 + \ldots + 3^{11}$$ is a G.P. with first term $$3$$, common ratio $$3$$ and $$11$$ terms:

$$\sum_{k=1}^{11} 3^{k} = \dfrac{3(3^{11} - 1)}{3 - 1} = \dfrac{3(3^{11} - 1)}{2}$$

Since $$3^{11} = 177147$$,

$$\sum_{k=1}^{11} 3^{k} = \dfrac{3(177147 - 1)}{2} = \dfrac{3 \times 177146}{2} = \dfrac{531438}{2} = 265719$$

Total:

$$\sum_{k=1}^{11} (2 + 3^k) = 22 + 265719 = 265741$$

Let the three terms of the G.P. be $$\dfrac{a}{r}, \ a, \ ar$$.

Using the product:

$$\dfrac{a}{r} \cdot a \cdot ar = a^3 = 1 \quad\Rightarrow\quad a = 1$$

Using the sum:

$$\dfrac{a}{r} + a + ar = \dfrac{39}{10}$$

Substituting $$a = 1$$:

$$\dfrac{1}{r} + 1 + r = \dfrac{39}{10} \quad\Rightarrow\quad \dfrac{1}{r} + r = \dfrac{39}{10} - 1 = \dfrac{29}{10}$$

Multiplying through by $$10r$$:

$$10 + 10r^2 = 29r \quad\Rightarrow\quad 10r^2 - 29r + 10 = 0$$

Solving this quadratic:

$$r = \dfrac{29 \pm \sqrt{841 - 400}}{20} = \dfrac{29 \pm \sqrt{441}}{20} = \dfrac{29 \pm 21}{20}$$

$$r = \dfrac{50}{20} = \dfrac{5}{2} \qquad \text{or} \qquad r = \dfrac{8}{20} = \dfrac{2}{5}$$

The terms. With $$a = 1$$ and $$r = \dfrac{5}{2}$$:

$$\dfrac{a}{r} = \dfrac{2}{5}, \qquad a = 1, \qquad ar = \dfrac{5}{2}$$

So the terms are $$\dfrac{2}{5}, 1, \dfrac{5}{2}$$. With $$r = \dfrac{2}{5}$$ the same numbers occur in reverse order: $$\dfrac{5}{2}, 1, \dfrac{2}{5}$$.

In the G.P. $$3, 3^2, 3^3, \ldots$$ the first term is $$a = 3$$ and the common ratio is $$r = 3$$.

The sum of $$n$$ terms is

$$S_n = \dfrac{a(r^{n} - 1)}{r - 1} = \dfrac{3(3^{n} - 1)}{3 - 1} = \dfrac{3(3^{n} - 1)}{2}$$

We are told this sum equals $$120$$:

$$\dfrac{3(3^{n} - 1)}{2} = 120$$

$$3(3^{n} - 1) = 240 \quad\Rightarrow\quad 3^{n} - 1 = 80 \quad\Rightarrow\quad 3^{n} = 81$$

Since $$81 = 3^{4}$$,

$$3^{n} = 3^{4} \quad\Rightarrow\quad n = 4$$

Hence $$4$$ terms are needed.

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

Sum of first three terms is $$16$$:

$$a + ar + ar^2 = 16 \quad\Rightarrow\quad a(1 + r + r^2) = 16 \qquad \text{...(1)}$$

Sum of next three terms is $$128$$: these are the 4th, 5th and 6th terms.

$$ar^3 + ar^4 + ar^5 = 128 \quad\Rightarrow\quad ar^3(1 + r + r^2) = 128 \qquad \text{...(2)}$$

Dividing (2) by (1):

$$\dfrac{ar^3(1 + r + r^2)}{a(1 + r + r^2)} = \dfrac{128}{16} \quad\Rightarrow\quad r^3 = 8 \quad\Rightarrow\quad r = 2$$

Substituting $$r = 2$$ in (1):

$$a(1 + 2 + 4) = 16 \quad\Rightarrow\quad 7a = 16 \quad\Rightarrow\quad a = \dfrac{16}{7}$$

Sum to $$n$$ terms:

$$S_n = \dfrac{a(r^{n} - 1)}{r - 1} = \dfrac{\frac{16}{7}(2^{n} - 1)}{2 - 1} = \dfrac{16}{7}(2^{n} - 1)$$

The first term is $$a = 729$$. The 7th term is $$64$$:

$$a r^{6} = 64 \quad\Rightarrow\quad 729\, r^{6} = 64$$

$$r^{6} = \dfrac{64}{729} = \dfrac{2^6}{3^6} = \left(\dfrac{2}{3}\right)^{6}$$

Taking the positive root, $$r = \dfrac{2}{3}$$.

The sum of $$7$$ terms, when the last term $$\ell$$ is known, is conveniently found from

$$S_n = \dfrac{a - \ell r}{1 - r}$$

Here $$\ell = 64$$ and $$r = \dfrac{2}{3}$$:

$$S_7 = \dfrac{729 - 64 \cdot \frac{2}{3}}{1 - \frac{2}{3}} = \dfrac{729 - \frac{128}{3}}{\frac{1}{3}}$$

$$S_7 = 3\left(729 - \dfrac{128}{3}\right) = 2187 - 128 = 2059$$

Hence $$S_7 = 2059$$.

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

Fifth term is $$4$$ times the third term:

$$a r^{4} = 4\,(a r^{2}) \quad\Rightarrow\quad r^{2} = 4 \quad\Rightarrow\quad r = \pm 2$$

Sum of first two terms is $$-4$$:

$$a + ar = -4 \quad\Rightarrow\quad a(1 + r) = -4 \qquad \text{...(1)}$$

Case 1: $$r = 2$$. From (1): $$a(1 + 2) = -4$$, so $$3a = -4$$ and $$a = -\dfrac{4}{3}$$.

The G.P. is $$-\dfrac{4}{3}, -\dfrac{8}{3}, -\dfrac{16}{3}, \ldots$$

Case 2: $$r = -2$$. From (1): $$a(1 - 2) = -4$$, so $$-a = -4$$ and $$a = 4$$.

The G.P. is $$4, -8, 16, -32, \ldots$$

Let the first term of the G.P. be $$a$$ and the common ratio be $$r$$. The general term is $$a_n = a r^{n-1}$$.

The 4th, 10th and 16th terms are

$$x = a r^{3}, \qquad y = a r^{9}, \qquad z = a r^{15}$$

Three numbers $$x, y, z$$ are in G.P. if $$y^2 = xz$$. We check this:

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

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

Since $$y^2 = xz = a^2 r^{18}$$, the numbers $$x, y, z$$ are in G.P.

(The common ratio of this new G.P. is $$\dfrac{y}{x} = r^{6}$$.) Hence proved.

This is not a G.P. We convert it to one. Let

$$S = 8 + 88 + 888 + \ldots \text{ to } n \text{ terms}$$

Take out the common factor $$8$$, then multiply and divide by $$9$$:

$$S = 8(1 + 11 + 111 + \ldots) = \dfrac{8}{9}(9 + 99 + 999 + \ldots \text{ to } n \text{ terms})$$

Write each term as a power of $$10$$ minus $$1$$:

$$S = \dfrac{8}{9}\big[(10 - 1) + (10^2 - 1) + \ldots + (10^n - 1)\big]$$

$$S = \dfrac{8}{9}\big[(10 + 10^2 + \ldots + 10^n) - n\big]$$

The bracketed G.P. has first term $$10$$, common ratio $$10$$ and $$n$$ terms:

$$10 + 10^2 + \ldots + 10^n = \dfrac{10(10^n - 1)}{10 - 1} = \dfrac{10(10^n - 1)}{9}$$

Therefore

$$S = \dfrac{8}{9}\left[\dfrac{10(10^n - 1)}{9} - n\right] = \dfrac{80}{81}(10^n - 1) - \dfrac{8n}{9}$$

Multiply the corresponding terms of the two sequences:

$$2 \times 128 = 256$$

$$4 \times 32 = 128$$

$$8 \times 8 = 64$$

$$16 \times 2 = 32$$

$$32 \times \dfrac{1}{2} = 16$$

So the new sequence of products is

$$256, 128, 64, 32, 16$$

This is a G.P. with first term $$a = 256$$, common ratio $$r = \dfrac{128}{256} = \dfrac{1}{2}$$, and $$n = 5$$ terms.

Its sum is

$$S_5 = \dfrac{a(1 - r^{5})}{1 - r} = \dfrac{256\left(1 - \left(\frac{1}{2}\right)^{5}\right)}{1 - \frac{1}{2}} = \dfrac{256\left(1 - \frac{1}{32}\right)}{\frac{1}{2}}$$

$$S_5 = 512 \times \dfrac{31}{32} = 16 \times 31 = 496$$

The two given sequences are G.P.s. The $$k$$th term of the first is $$a r^{k-1}$$ and the $$k$$th term of the second is $$A R^{k-1}$$.

Multiplying corresponding terms, the $$k$$th term of the product sequence is

$$T_k = (a r^{k-1})(A R^{k-1}) = aA\,(rR)^{k-1}$$

So the product sequence is

$$aA, \ aA(rR), \ aA(rR)^2, \ \ldots, \ aA(rR)^{n-1}$$

To confirm it is a G.P., take the ratio of any term to the previous one:

$$\dfrac{T_{k+1}}{T_k} = \dfrac{aA\,(rR)^{k}}{aA\,(rR)^{k-1}} = rR$$

This ratio is constant (independent of $$k$$), so the product sequence is a G.P. with first term $$aA$$ and common ratio $$rR$$.

Let the four numbers in G.P. be $$a, \ ar, \ ar^2, \ ar^3$$.

Third term exceeds the first term by $$9$$:

$$ar^2 - a = 9 \quad\Rightarrow\quad a(r^2 - 1) = 9 \qquad \text{...(1)}$$

Second term exceeds the fourth term by $$18$$:

$$ar - ar^3 = 18 \quad\Rightarrow\quad ar(1 - r^2) = 18 \qquad \text{...(2)}$$

Rewrite (2) as $$-ar(r^2 - 1) = 18$$. Dividing this by (1):

$$\dfrac{-ar(r^2 - 1)}{a(r^2 - 1)} = \dfrac{18}{9} \quad\Rightarrow\quad -r = 2 \quad\Rightarrow\quad r = -2$$

Substituting $$r = -2$$ in (1):

$$a\big((-2)^2 - 1\big) = 9 \quad\Rightarrow\quad a(4 - 1) = 9 \quad\Rightarrow\quad 3a = 9 \quad\Rightarrow\quad a = 3$$

The four numbers are

$$a = 3, \quad ar = -6, \quad ar^2 = 12, \quad ar^3 = -24$$

Hence the four numbers are $$3, -6, 12, -24$$.

Let the first term of the G.P. be $$A$$ and the common ratio be $$R$$. Then the $$p$$th, $$q$$th and $$r$$th terms are

$$a = A R^{p-1}, \qquad b = A R^{q-1}, \qquad c = A R^{r-1}$$

Substitute these into the expression $$a^{q-r}\, b^{r-p}\, c^{p-q}$$:

$$a^{q-r} = A^{q-r}\, R^{(p-1)(q-r)}$$

$$b^{r-p} = A^{r-p}\, R^{(q-1)(r-p)}$$

$$c^{p-q} = A^{p-q}\, R^{(r-1)(p-q)}$$

Multiplying, the power of $$A$$ is

$$(q-r) + (r-p) + (p-q) = 0$$

so $$A^{0} = 1$$.

The power of $$R$$ is

$$(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)$$

Expand by separating each product:

$$= \big[p(q-r) + q(r-p) + r(p-q)\big] - \big[(q-r) + (r-p) + (p-q)\big]$$

The first bracket is $$pq - pr + qr - pq + rp - rq = 0$$, and the second bracket is also $$0$$. So the total power of $$R$$ is $$0$$, giving $$R^{0} = 1$$.

Therefore

$$a^{q-r}\, b^{r-p}\, c^{p-q} = A^{0} \cdot R^{0} = 1 \times 1 = 1$$

Hence proved.

Let the common ratio be $$r$$. The first term is $$a$$, so the $$n$$ terms are

$$a, \ ar, \ ar^2, \ \ldots, \ ar^{n-1}$$

The $$n$$th term is $$b = a r^{n-1}$$.

The product of all $$n$$ terms is

$$P = a \cdot ar \cdot ar^2 \cdots ar^{n-1} = a^{n}\, r^{\,0 + 1 + 2 + \ldots + (n-1)}$$

The exponent of $$r$$ is the sum of the first $$(n-1)$$ whole numbers:

$$0 + 1 + 2 + \ldots + (n-1) = \dfrac{n(n-1)}{2}$$

So

$$P = a^{n}\, r^{\frac{n(n-1)}{2}}$$

Squaring:

$$P^2 = a^{2n}\, r^{n(n-1)} \qquad \text{...(1)}$$

Now compute $$(ab)^n$$. Since $$b = a r^{n-1}$$,

$$ab = a \cdot a r^{n-1} = a^2 r^{n-1}$$

$$(ab)^n = (a^2 r^{n-1})^{n} = a^{2n}\, r^{n(n-1)} \qquad \text{...(2)}$$

From (1) and (2), $$P^2 = (ab)^n$$. Hence proved.

Let the G.P. have first term $$a$$ and common ratio $$r$$.

Sum of the first $$n$$ terms:

$$S_n = \dfrac{a(r^{n} - 1)}{r - 1}$$

Sum of the terms from the $$(n+1)$$th to the $$(2n)$$th: these are $$n$$ consecutive terms of the G.P. The first of them is the $$(n+1)$$th term, $$a r^{n}$$, and the common ratio is still $$r$$. Hence their sum is

$$S' = \dfrac{a r^{n}(r^{n} - 1)}{r - 1}$$

(One can also see this as $$S_{2n} - S_n$$, which gives the same result.)

The required ratio:

$$\dfrac{S_n}{S'} = \dfrac{\dfrac{a(r^{n} - 1)}{r - 1}}{\dfrac{a r^{n}(r^{n} - 1)}{r - 1}}$$

Cancelling the common factor $$\dfrac{a(r^{n} - 1)}{r - 1}$$:

$$\dfrac{S_n}{S'} = \dfrac{1}{r^{n}}$$

Hence proved.

Since $$a, b, c, d$$ are in G.P., write each in terms of the first term $$a$$ and the common ratio $$r$$:

$$b = ar, \qquad c = ar^2, \qquad d = ar^3$$

Left-hand side. First factor:

$$a^2 + b^2 + c^2 = a^2 + a^2 r^2 + a^2 r^4 = a^2(1 + r^2 + r^4)$$

Second factor:

$$b^2 + c^2 + d^2 = a^2 r^2 + a^2 r^4 + a^2 r^6 = a^2 r^2(1 + r^2 + r^4)$$

Multiplying the two factors:

$$(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = a^2(1 + r^2 + r^4) \cdot a^2 r^2(1 + r^2 + r^4) = a^4 r^2 (1 + r^2 + r^4)^2$$

Right-hand side. Compute each product:

$$ab = a \cdot ar = a^2 r, \qquad bc = ar \cdot ar^2 = a^2 r^3, \qquad cd = ar^2 \cdot ar^3 = a^2 r^5$$

So

$$ab + bc + cd = a^2 r + a^2 r^3 + a^2 r^5 = a^2 r(1 + r^2 + r^4)$$

Squaring:

$$(ab + bc + cd)^2 = a^4 r^2 (1 + r^2 + r^4)^2$$

Since the LHS and RHS are both equal to $$a^4 r^2 (1 + r^2 + r^4)^2$$,

$$(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2$$

Hence proved.

Let the two numbers inserted be $$G_1$$ and $$G_2$$. Then

$$3, \ G_1, \ G_2, \ 81$$

is a G.P. with $$4$$ terms, first term $$a = 3$$ and 4th term $$81$$.

Using $$a_4 = a r^{3}$$:

$$3\, r^{3} = 81 \quad\Rightarrow\quad r^{3} = 27 \quad\Rightarrow\quad r = 3$$

Therefore

$$G_1 = ar = 3 \times 3 = 9, \qquad G_2 = ar^2 = 3 \times 9 = 27$$

Hence the two numbers are $$9$$ and $$27$$, and the G.P. is $$3, 9, 27, 81$$.

The geometric mean between $$a$$ and $$b$$ is $$\sqrt{ab} = a^{1/2} b^{1/2}$$. So we require

$$\dfrac{a^{n+1} + b^{n+1}}{a^{n} + b^{n}} = a^{1/2} b^{1/2}$$

Cross-multiplying:

$$a^{n+1} + b^{n+1} = a^{1/2} b^{1/2}\,(a^{n} + b^{n}) = a^{n + \frac{1}{2}} b^{\frac{1}{2}} + a^{\frac{1}{2}} b^{n + \frac{1}{2}}$$

Bring like terms together:

$$a^{n+1} - a^{n + \frac{1}{2}} b^{\frac{1}{2}} = a^{\frac{1}{2}} b^{n + \frac{1}{2}} - b^{n+1}$$

Factor each side:

$$a^{n + \frac{1}{2}}\left(a^{\frac{1}{2}} - b^{\frac{1}{2}}\right) = b^{n + \frac{1}{2}}\left(a^{\frac{1}{2}} - b^{\frac{1}{2}}\right)$$

Since $$a \neq b$$, the factor $$\left(a^{1/2} - b^{1/2}\right) \neq 0$$, so we may cancel it:

$$a^{n + \frac{1}{2}} = b^{n + \frac{1}{2}} \quad\Rightarrow\quad \left(\dfrac{a}{b}\right)^{n + \frac{1}{2}} = 1$$

As $$a \neq b$$, the ratio $$\dfrac{a}{b} \neq 1$$, so the exponent must be zero:

$$n + \dfrac{1}{2} = 0 \quad\Rightarrow\quad n = -\dfrac{1}{2}$$

Let the two positive numbers be $$a$$ and $$b$$. Their geometric mean is $$\sqrt{ab}$$. We are given

$$a + b = 6\sqrt{ab} \qquad \text{...(1)}$$

We use the componendo-dividendo technique. From (1), consider

$$\dfrac{a + b + 2\sqrt{ab}}{a + b - 2\sqrt{ab}}$$

Replacing $$a + b$$ by $$6\sqrt{ab}$$ in numerator and denominator:

$$\dfrac{a + b + 2\sqrt{ab}}{a + b - 2\sqrt{ab}} = \dfrac{6\sqrt{ab} + 2\sqrt{ab}}{6\sqrt{ab} - 2\sqrt{ab}} = \dfrac{8\sqrt{ab}}{4\sqrt{ab}} = 2$$

But $$a + b + 2\sqrt{ab} = (\sqrt{a} + \sqrt{b})^2$$ and $$a + b - 2\sqrt{ab} = (\sqrt{a} - \sqrt{b})^2$$. So

$$\dfrac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = 2 \quad\Rightarrow\quad \dfrac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \sqrt{2}$$

Apply componendo-dividendo again (add and subtract numerator and denominator):

$$\dfrac{(\sqrt{a} + \sqrt{b}) + (\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b}) - (\sqrt{a} - \sqrt{b})} = \dfrac{\sqrt{2} + 1}{\sqrt{2} - 1}$$

$$\dfrac{2\sqrt{a}}{2\sqrt{b}} = \dfrac{\sqrt{2} + 1}{\sqrt{2} - 1} \quad\Rightarrow\quad \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{2} + 1}{\sqrt{2} - 1}$$

Squaring both sides:

$$\dfrac{a}{b} = \dfrac{(\sqrt{2} + 1)^2}{(\sqrt{2} - 1)^2} = \dfrac{2 + 2\sqrt{2} + 1}{2 - 2\sqrt{2} + 1} = \dfrac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$$

Hence $$a : b = (3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$$. Proved.

Let the two positive numbers be $$a$$ and $$b$$.

Their A.M. is $$A = \dfrac{a + b}{2}$$, which gives

$$a + b = 2A \qquad \text{...(1)}$$

Their G.M. is $$G = \sqrt{ab}$$, which gives

$$ab = G^2 \qquad \text{...(2)}$$

We find $$a - b$$ using the identity $$(a - b)^2 = (a + b)^2 - 4ab$$:

$$(a - b)^2 = (2A)^2 - 4G^2 = 4A^2 - 4G^2 = 4(A^2 - G^2)$$

Factor $$A^2 - G^2 = (A + G)(A - G)$$:

$$(a - b)^2 = 4(A + G)(A - G)$$

$$a - b = \pm\, 2\sqrt{(A + G)(A - G)} \qquad \text{...(3)}$$

Adding (1) and (3) and dividing by $$2$$:

$$a = \dfrac{(a + b) + (a - b)}{2} = \dfrac{2A \pm 2\sqrt{(A + G)(A - G)}}{2} = A \pm \sqrt{(A + G)(A - G)}$$

The other number $$b$$ takes the remaining sign. Hence the two numbers are

$$A \pm \sqrt{(A + G)(A - G)}$$

Hence proved.

Originally there are $$30$$ bacteria. Since the count doubles every hour, the numbers at the end of each hour form a G.P. with common ratio $$r = 2$$:

End of 1st hour: $$30 \times 2 = 60$$

End of 2nd hour: $$60 \times 2 = 120$$, which is $$30 \times 2^2$$

End of 3rd hour: $$30 \times 2^3$$, and so on.

So the number of bacteria at the end of the $$n$$th hour is

$$30 \times 2^{n}$$

End of 2nd hour ($$n = 2$$):

$$30 \times 2^{2} = 30 \times 4 = 120$$

End of 4th hour ($$n = 4$$):

$$30 \times 2^{4} = 30 \times 16 = 480$$

End of $$n$$th hour: $$30 \times 2^{n}$$.

The interest is compounded annually at $$10\%$$. Each year the amount is multiplied by

$$1 + \dfrac{10}{100} = 1.1$$

So the amounts at the end of successive years form a G.P.:

End of year 1: $$500(1.1)$$, end of year 2: $$500(1.1)^2$$, and so on.

Hence the amount at the end of $$10$$ years is

$$A = 500\,(1.1)^{10}$$

Evaluating $$(1.1)^{10} \approx 2.5937$$,

$$A \approx 500 \times 2.5937 = 1296.87$$

So Rs $$500$$ amounts to about Rs $$1296.87$$ in $$10$$ years.

Let the roots of the required quadratic equation be $$\alpha$$ and $$\beta$$.

A.M. of the roots is $$8$$:

$$\dfrac{\alpha + \beta}{2} = 8 \quad\Rightarrow\quad \alpha + \beta = 16$$

G.M. of the roots is $$5$$:

$$\sqrt{\alpha\beta} = 5 \quad\Rightarrow\quad \alpha\beta = 25$$

A quadratic equation with roots $$\alpha$$ and $$\beta$$ is

$$x^2 - (\alpha + \beta)x + \alpha\beta = 0$$

Substituting the values:

$$x^2 - 16x + 25 = 0$$

Consider the given expression and try to write it as a sum of perfect squares.

Expand $$(ap - b)^2 + (bp - c)^2 + (cp - d)^2$$:

$$(ap - b)^2 = a^2p^2 - 2abp + b^2$$

$$(bp - c)^2 = b^2p^2 - 2bcp + c^2$$

$$(cp - d)^2 = c^2p^2 - 2cdp + d^2$$

Adding these three results and grouping like terms:

$$(ap - b)^2 + (bp - c)^2 + (cp - d)^2 = (a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2)$$

So the given expression is exactly this sum of squares. We are told

$$(ap - b)^2 + (bp - c)^2 + (cp - d)^2 \leq 0$$

But a sum of squares of real numbers can never be negative; it is always $$\geq 0$$. The only way it can also be $$\leq 0$$ is if it equals $$0$$:

$$(ap - b)^2 + (bp - c)^2 + (cp - d)^2 = 0$$

A sum of squares is zero only when each square is zero individually:

$$ap - b = 0, \qquad bp - c = 0, \qquad cp - d = 0$$

That is,

$$\dfrac{b}{a} = p, \qquad \dfrac{c}{b} = p, \qquad \dfrac{d}{c} = p$$

Since the ratio of each term to the previous one is the same constant $$p$$, the numbers $$a, b, c, d$$ are in G.P. (with common ratio $$p$$). Hence proved.

We are given $$f(x+y) = f(x)\,f(y)$$ and $$f(1) = 3$$. Use the relation to find the values of $$f$$ at $$1, 2, 3, \ldots$$

$$f(2) = f(1+1) = f(1)\,f(1) = 3 \times 3 = 9 = 3^2$$

$$f(3) = f(2+1) = f(2)\,f(1) = 9 \times 3 = 27 = 3^3$$

In general, $$f(x) = 3^{x}$$.

So the sum becomes a G.P. with first term $$3$$, common ratio $$3$$ and $$n$$ terms:

$$\sum_{x=1}^{n} f(x) = 3 + 3^2 + 3^3 + \ldots + 3^{n} = \dfrac{3(3^{n} - 1)}{3 - 1} = \dfrac{3(3^{n} - 1)}{2}$$

Setting this equal to $$120$$:

$$\dfrac{3(3^{n} - 1)}{2} = 120 \quad\Rightarrow\quad 3(3^{n} - 1) = 240 \quad\Rightarrow\quad 3^{n} - 1 = 80$$

$$3^{n} = 81 = 3^{4} \quad\Rightarrow\quad n = 4$$

The G.P. has first term $$a = 5$$ and common ratio $$r = 2$$. Let the number of terms be $$n$$.

The sum of $$n$$ terms is

$$S_n = \dfrac{a(r^{n} - 1)}{r - 1} = \dfrac{5(2^{n} - 1)}{2 - 1} = 5(2^{n} - 1)$$

Setting $$S_n = 315$$:

$$5(2^{n} - 1) = 315 \quad\Rightarrow\quad 2^{n} - 1 = 63 \quad\Rightarrow\quad 2^{n} = 64$$

Since $$64 = 2^{6}$$, we get $$n = 6$$.

The last term is the $$n$$th term:

$$a_n = a r^{n-1} = 5 \times 2^{5} = 5 \times 32 = 160$$

Hence there are $$6$$ terms and the last term is $$160$$.

The first term is $$a = 1$$. Let the common ratio be $$r$$. The general term is $$a_n = a r^{n-1}$$.

Third term: $$a_3 = a r^2 = r^2$$.

Fifth term: $$a_5 = a r^4 = r^4$$.

Their sum is $$90$$:

$$r^2 + r^4 = 90$$

Put $$y = r^2$$ (so $$y \geq 0$$):

$$y + y^2 = 90 \quad\Rightarrow\quad y^2 + y - 90 = 0$$

Solve the quadratic:

$$y = \dfrac{-1 \pm \sqrt{1 + 360}}{2} = \dfrac{-1 \pm \sqrt{361}}{2} = \dfrac{-1 \pm 19}{2}$$

$$y = 9 \qquad \text{or} \qquad y = -10$$

Since $$y = r^2$$ cannot be negative, we reject $$y = -10$$. Thus $$r^2 = 9$$, so

$$r = \pm 3$$

Let the three numbers in G.P. be $$a, \ ar, \ ar^2$$.

Their sum is $$56$$:

$$a + ar + ar^2 = 56 \qquad \text{...(1)}$$

After subtracting $$1, 7, 21$$ the numbers $$a - 1, \ ar - 7, \ ar^2 - 21$$ are in A.P. For an A.P., twice the middle term equals the sum of the outer terms:

$$2(ar - 7) = (a - 1) + (ar^2 - 21)$$

$$2ar - 14 = a + ar^2 - 22$$

$$2ar = a + ar^2 - 8 \qquad \text{...(2)}$$

From (1), $$a + ar^2 = 56 - ar$$. Substitute into (2):

$$2ar = (56 - ar) - 8 = 48 - ar$$

$$3ar = 48 \quad\Rightarrow\quad ar = 16$$

So the middle term is $$ar = 16$$. From (1), $$a + ar^2 = 56 - 16 = 40$$.

Also $$a \cdot ar^2 = (ar)^2 = 16^2 = 256$$.

Thus $$a$$ and $$ar^2$$ are two numbers with sum $$40$$ and product $$256$$, so they are roots of

$$t^2 - 40t + 256 = 0$$

$$t = \dfrac{40 \pm \sqrt{1600 - 1024}}{2} = \dfrac{40 \pm \sqrt{576}}{2} = \dfrac{40 \pm 24}{2}$$

$$t = 32 \qquad \text{or} \qquad t = 8$$

Taking $$a = 8$$ and $$ar^2 = 32$$ (with $$ar = 16$$), the three numbers are $$8, 16, 32$$. Taking $$a = 32$$ instead gives the same numbers in reverse order $$32, 16, 8$$.

Let the G.P. have $$2n$$ terms (an even number), with first term $$a$$ and common ratio $$r$$. The terms are

$$a, \ ar, \ ar^2, \ \ldots, \ ar^{2n-1}$$

Sum of all $$2n$$ terms:

$$S = \dfrac{a(r^{2n} - 1)}{r - 1}$$

Terms in odd places are the 1st, 3rd, 5th, $$\ldots$$ terms: $$a, ar^2, ar^4, \ldots, ar^{2n-2}$$. These form a G.P. with first term $$a$$, common ratio $$r^2$$ and $$n$$ terms:

$$S_{\text{odd}} = \dfrac{a\big((r^2)^{n} - 1\big)}{r^2 - 1} = \dfrac{a(r^{2n} - 1)}{r^2 - 1}$$

Given condition: $$S = 5\,S_{\text{odd}}$$.

$$\dfrac{a(r^{2n} - 1)}{r - 1} = 5 \cdot \dfrac{a(r^{2n} - 1)}{r^2 - 1}$$

Cancel the common factor $$a(r^{2n} - 1)$$ (which is non-zero):

$$\dfrac{1}{r - 1} = \dfrac{5}{r^2 - 1} = \dfrac{5}{(r - 1)(r + 1)}$$

Multiply both sides by $$(r - 1)$$:

$$1 = \dfrac{5}{r + 1} \quad\Rightarrow\quad r + 1 = 5 \quad\Rightarrow\quad r = 4$$

Hence the common ratio is $$4$$.

We are given

$$\dfrac{a+bx}{a-bx} = \dfrac{b+cx}{b-cx} = \dfrac{c+dx}{c-dx}$$

Take the first two equal fractions and apply componendo-dividendo (add and subtract numerator and denominator):

$$\dfrac{(a+bx) + (a-bx)}{(a+bx) - (a-bx)} = \dfrac{(b+cx) + (b-cx)}{(b+cx) - (b-cx)}$$

$$\dfrac{2a}{2bx} = \dfrac{2b}{2cx} \quad\Rightarrow\quad \dfrac{a}{bx} = \dfrac{b}{cx}$$

Since $$x \neq 0$$, multiply both sides by $$x$$:

$$\dfrac{a}{b} = \dfrac{b}{c} \quad\Rightarrow\quad b^2 = ac \qquad \text{...(1)}$$

Take the last two equal fractions and apply componendo-dividendo similarly:

$$\dfrac{(b+cx) + (b-cx)}{(b+cx) - (b-cx)} = \dfrac{(c+dx) + (c-dx)}{(c+dx) - (c-dx)}$$

$$\dfrac{2b}{2cx} = \dfrac{2c}{2dx} \quad\Rightarrow\quad \dfrac{b}{c} = \dfrac{c}{d} \quad\Rightarrow\quad c^2 = bd \qquad \text{...(2)}$$

From (1) and (2):

$$\dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c}$$

The ratio of each term to the previous one is the same, so $$a, b, c, d$$ are in G.P. Hence proved.

Let the G.P. have first term $$a$$ and common ratio $$r$$. Its $$n$$ terms are

$$a, \ ar, \ ar^2, \ \ldots, \ ar^{n-1}$$

Sum $$S$$:

$$S = \dfrac{a(r^{n} - 1)}{r - 1}$$

Product $$P$$:

$$P = a \cdot ar \cdot ar^2 \cdots ar^{n-1} = a^{n}\, r^{0 + 1 + \ldots + (n-1)} = a^{n}\, r^{\frac{n(n-1)}{2}}$$

Sum of reciprocals $$R$$: the reciprocals are $$\dfrac{1}{a}, \dfrac{1}{ar}, \ldots, \dfrac{1}{ar^{n-1}}$$, themselves a G.P. with first term $$\dfrac{1}{a}$$ and common ratio $$\dfrac{1}{r}$$:

$$R = \dfrac{\frac{1}{a}\left(1 - \frac{1}{r^{n}}\right)}{1 - \frac{1}{r}} = \dfrac{\frac{1}{a}\cdot\frac{r^{n} - 1}{r^{n}}}{\frac{r - 1}{r}} = \dfrac{r^{n} - 1}{a\, r^{n-1}(r - 1)}$$

Now form $$\dfrac{S}{R}$$:

$$\dfrac{S}{R} = \dfrac{\dfrac{a(r^{n} - 1)}{r - 1}}{\dfrac{r^{n} - 1}{a\, r^{n-1}(r - 1)}} = \dfrac{a(r^{n} - 1)}{r - 1} \cdot \dfrac{a\, r^{n-1}(r - 1)}{r^{n} - 1} = a^2 r^{n-1}$$

Raising to the $$n$$th power:

$$\left(\dfrac{S}{R}\right)^{n} = (a^2 r^{n-1})^{n} = a^{2n}\, r^{n(n-1)}$$

On the other hand,

$$P^2 = \left(a^{n}\, r^{\frac{n(n-1)}{2}}\right)^2 = a^{2n}\, r^{n(n-1)}$$

Therefore $$P^2 = \left(\dfrac{S}{R}\right)^{n} = \dfrac{S^{n}}{R^{n}}$$, which gives

$$P^2 R^{n} = S^{n}$$

Hence proved.

Since $$a, b, c, d$$ are in G.P., let the common ratio be $$k$$. Then

$$b = ak, \qquad c = ak^2, \qquad d = ak^3$$

Compute each of the three given expressions.

$$a^n + b^n = a^n + (ak)^n = a^n + a^n k^n = a^n(1 + k^n)$$

$$b^n + c^n = (ak)^n + (ak^2)^n = a^n k^n + a^n k^{2n} = a^n k^n(1 + k^n)$$

$$c^n + d^n = (ak^2)^n + (ak^3)^n = a^n k^{2n} + a^n k^{3n} = a^n k^{2n}(1 + k^n)$$

For these to be in G.P., the ratio of consecutive terms must be equal. Compute the two ratios:

$$\dfrac{b^n + c^n}{a^n + b^n} = \dfrac{a^n k^n(1 + k^n)}{a^n(1 + k^n)} = k^n$$

$$\dfrac{c^n + d^n}{b^n + c^n} = \dfrac{a^n k^{2n}(1 + k^n)}{a^n k^n(1 + k^n)} = k^n$$

Both ratios equal $$k^n$$. Therefore $$(a^n + b^n), (b^n + c^n), (c^n + d^n)$$ are in G.P. (with common ratio $$k^n$$). Hence proved.

Since $$a$$ and $$b$$ are roots of $$x^2 - 3x + p = 0$$, by the sum and product of roots:

$$a + b = 3, \qquad ab = p$$

Since $$c$$ and $$d$$ are roots of $$x^2 - 12x + q = 0$$:

$$c + d = 12, \qquad cd = q$$

As $$a, b, c, d$$ are in G.P., let the first term be $$a$$ and common ratio $$r$$, so

$$b = ar, \qquad c = ar^2, \qquad d = ar^3$$

Using the sums:

$$a + b = a + ar = a(1 + r) = 3 \qquad \text{...(1)}$$

$$c + d = ar^2 + ar^3 = ar^2(1 + r) = 12 \qquad \text{...(2)}$$

Dividing (2) by (1):

$$\dfrac{ar^2(1 + r)}{a(1 + r)} = \dfrac{12}{3} \quad\Rightarrow\quad r^2 = 4$$

So $$r^2 = 4$$, and hence $$r^4 = (r^2)^2 = 16$$.

Using the products:

$$p = ab = a \cdot ar = a^2 r$$

$$q = cd = ar^2 \cdot ar^3 = a^2 r^5$$

Now form the required ratio:

$$\dfrac{q + p}{q - p} = \dfrac{a^2 r^5 + a^2 r}{a^2 r^5 - a^2 r} = \dfrac{a^2 r(r^4 + 1)}{a^2 r(r^4 - 1)} = \dfrac{r^4 + 1}{r^4 - 1}$$

Substituting $$r^4 = 16$$:

$$\dfrac{q + p}{q - p} = \dfrac{16 + 1}{16 - 1} = \dfrac{17}{15}$$

Hence $$(q + p) : (q - p) = 17 : 15$$. Proved.

For the two positive numbers $$a$$ and $$b$$, the A.M. is $$\dfrac{a+b}{2}$$ and the G.M. is $$\sqrt{ab}$$. Their ratio is $$m : n$$:

$$\dfrac{\frac{a+b}{2}}{\sqrt{ab}} = \dfrac{m}{n} \quad\Rightarrow\quad \dfrac{a+b}{2\sqrt{ab}} = \dfrac{m}{n}$$

So

$$a + b = \dfrac{2m}{n}\sqrt{ab} \qquad \text{...(1)}$$

Form the expression with componendo-dividendo, using $$a + b \pm 2\sqrt{ab} = (\sqrt{a} \pm \sqrt{b})^2$$:

$$\dfrac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = \dfrac{a + b + 2\sqrt{ab}}{a + b - 2\sqrt{ab}}$$

Substitute $$a + b = \dfrac{2m}{n}\sqrt{ab}$$ from (1):

$$= \dfrac{\frac{2m}{n}\sqrt{ab} + 2\sqrt{ab}}{\frac{2m}{n}\sqrt{ab} - 2\sqrt{ab}} = \dfrac{\frac{2m}{n} + 2}{\frac{2m}{n} - 2} = \dfrac{m + n}{m - n}$$

Therefore

$$\dfrac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \sqrt{\dfrac{m + n}{m - n}} = \dfrac{\sqrt{m + n}}{\sqrt{m - n}}$$

Apply componendo-dividendo once more:

$$\dfrac{(\sqrt{a} + \sqrt{b}) + (\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b}) - (\sqrt{a} - \sqrt{b})} = \dfrac{\sqrt{m + n} + \sqrt{m - n}}{\sqrt{m + n} - \sqrt{m - n}}$$

$$\dfrac{2\sqrt{a}}{2\sqrt{b}} = \dfrac{\sqrt{m + n} + \sqrt{m - n}}{\sqrt{m + n} - \sqrt{m - n}} \quad\Rightarrow\quad \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{m + n} + \sqrt{m - n}}{\sqrt{m + n} - \sqrt{m - n}}$$

Squaring both sides:

$$\dfrac{a}{b} = \dfrac{(\sqrt{m+n} + \sqrt{m-n})^2}{(\sqrt{m+n} - \sqrt{m-n})^2} = \dfrac{(m+n) + (m-n) + 2\sqrt{(m+n)(m-n)}}{(m+n) + (m-n) - 2\sqrt{(m+n)(m-n)}}$$

$$\dfrac{a}{b} = \dfrac{2m + 2\sqrt{m^2 - n^2}}{2m - 2\sqrt{m^2 - n^2}} = \dfrac{m + \sqrt{m^2 - n^2}}{m - \sqrt{m^2 - n^2}}$$

Hence $$a : b = \left(m + \sqrt{m^2 - n^2}\right) : \left(m - \sqrt{m^2 - n^2}\right)$$. Proved.

Look at the pattern of the terms:

1st term: $$2 \times 4$$, 2nd term: $$4 \times 6$$, 3rd term: $$6 \times 8$$, $$\ldots$$

In the $$n$$th term, the first factor is $$2n$$ (the numbers $$2, 4, 6, \ldots$$) and the second factor is $$2n + 2$$ (the numbers $$4, 6, 8, \ldots$$). So the $$n$$th term is

$$a_n = (2n)(2n + 2)$$

For the 20th term, put $$n = 20$$:

$$a_{20} = (2 \times 20)(2 \times 20 + 2) = (40)(42) = 1680$$

Hence the 20th term is $$1680$$.

The tractor costs Rs $$12000$$. The farmer pays Rs $$6000$$ cash, so the balance to be paid in instalments is

$$12000 - 6000 = 6000$$

This balance is cleared by yearly instalments of Rs $$500$$, so the number of instalments is

$$\dfrac{6000}{500} = 12$$

Each year, besides the instalment, he pays $$12\%$$ interest on the amount still unpaid. The unpaid amounts at the start of each year are

$$6000, \ 5500, \ 5000, \ \ldots, \ 500$$

The total interest is $$12\%$$ of the sum of all these unpaid amounts:

$$\text{Total interest} = \dfrac{12}{100}\,(6000 + 5500 + 5000 + \ldots + 500)$$

The sum in brackets is an A.P. with first term $$6000$$, last term $$500$$ and $$12$$ terms:

$$6000 + 5500 + \ldots + 500 = \dfrac{12}{2}(6000 + 500) = 6 \times 6500 = 39000$$

So

$$\text{Total interest} = \dfrac{12}{100} \times 39000 = 4680$$

The total cost of the tractor is the marked price plus the total interest:

$$12000 + 4680 = 16680$$

Hence the tractor cost him Rs $$16680$$.

The scooter costs Rs $$22000$$. He pays Rs $$4000$$ cash, so the balance to be paid in instalments is

$$22000 - 4000 = 18000$$

This balance is cleared by yearly instalments of Rs $$1000$$, so the number of instalments is

$$\dfrac{18000}{1000} = 18$$

Each year he also pays $$10\%$$ interest on the unpaid amount. The unpaid amounts at the start of each year are

$$18000, \ 17000, \ 16000, \ \ldots, \ 1000$$

The total interest is $$10\%$$ of the sum of all these amounts:

$$\text{Total interest} = \dfrac{10}{100}\,(18000 + 17000 + \ldots + 1000)$$

The sum in brackets is an A.P. with first term $$18000$$, last term $$1000$$ and $$18$$ terms:

$$18000 + 17000 + \ldots + 1000 = \dfrac{18}{2}(18000 + 1000) = 9 \times 19000 = 171000$$

So

$$\text{Total interest} = \dfrac{10}{100} \times 171000 = 17100$$

The total cost of the scooter is the marked price plus the total interest:

$$22000 + 17100 = 39100$$

Hence the scooter will cost him Rs $$39100$$.

Count the letters mailed in each "set".

1st set: the person mails $$4$$ letters.

2nd set: each of those $$4$$ friends mails $$4$$ letters, giving $$4 \times 4 = 4^2 = 16$$ letters.

3rd set: $$4^3$$ letters, and so on.

So the numbers of letters in the successive sets form a G.P.:

$$4, \ 4^2, \ 4^3, \ \ldots, \ 4^8$$

The total number of letters mailed up to and including the 8th set is the sum of this G.P. with first term $$a = 4$$, common ratio $$r = 4$$ and $$n = 8$$ terms:

$$\text{Total letters} = \dfrac{a(r^{n} - 1)}{r - 1} = \dfrac{4(4^{8} - 1)}{4 - 1}$$

Since $$4^{8} = 65536$$,

$$\text{Total letters} = \dfrac{4(65536 - 1)}{3} = \dfrac{4 \times 65535}{3} = \dfrac{262140}{3} = 87380$$

Each letter costs $$50$$ paise $$= $$ Rs $$0.50$$ to mail. So the amount spent on postage is

$$87380 \times 0.50 = 43690$$

Hence the amount spent on postage is Rs $$43690$$.

Principal $$P = $$ Rs $$10\,000$$; rate $$R = 5\%$$ per annum simple interest.

Interest accrued every year is the same:

$$\text{Interest per year} = \dfrac{P \times R}{100} = \dfrac{10\,000 \times 5}{100} = 500.$$

So each year Rs $$500$$ of interest is added; the year-by-year amounts form an A.P. with first term equal to the amount after year 1 and common difference $$500$$.

Amount at the end of the 15th year. After $$15$$ years of interest have been added to the principal:

$$A_{15} = P + 15 \times 500 = 10\,000 + 7\,500 = 17\,500.$$

Total amount after 20 years. After $$20$$ years of interest have been added:

$$A_{20} = P + 20 \times 500 = 10\,000 + 10\,000 = 20\,000.$$

Hence the amount at the end of the 15th year is Rs $$17\,500$$, and the total amount after 20 years is Rs $$20\,000$$.

The machine costs Rs $$15625$$. Each year its value depreciates by $$20\%$$, so at the end of each year the value is

$$100\% - 20\% = 80\% = \dfrac{80}{100} = \dfrac{4}{5}$$

of the value at the start of that year. So the values at the end of successive years form a G.P. with common ratio $$\dfrac{4}{5}$$.

The value at the end of $$5$$ years is

$$\text{Value} = 15625 \times \left(\dfrac{4}{5}\right)^{5}$$

Compute $$\left(\dfrac{4}{5}\right)^{5} = \dfrac{4^5}{5^5} = \dfrac{1024}{3125}$$. Therefore

$$\text{Value} = 15625 \times \dfrac{1024}{3125} = \dfrac{15625}{3125} \times 1024 = 5 \times 1024 = 5120$$

Hence the estimated value of the machine at the end of $$5$$ years is Rs $$5120$$.

Suppose the work was originally planned to be finished in $$n$$ days. With $$150$$ workers working all $$n$$ days, the total work is

$$\text{Total work} = 150n \ \text{worker-days}$$

In reality, workers keep dropping out: $$150$$ work on day 1, $$146$$ on day 2, $$142$$ on day 3, and so on (decreasing by $$4$$ each day). Because of this, the job took $$8$$ extra days, i.e. $$(n + 8)$$ days in all.

The work actually done is the sum of an A.P. with first term $$150$$, common difference $$-4$$ and $$(n + 8)$$ terms:

$$\text{Work done} = \dfrac{n + 8}{2}\big[2(150) + (n + 8 - 1)(-4)\big]$$

$$= \dfrac{n + 8}{2}\big[300 - 4(n + 7)\big] = \dfrac{n + 8}{2}\big[300 - 4n - 28\big] = \dfrac{n + 8}{2}(272 - 4n)$$

Since the same job is completed, set the work done equal to the planned total work $$150n$$:

$$\dfrac{n + 8}{2}(272 - 4n) = 150n$$

$$(n + 8)(272 - 4n) = 300n$$

Expand the left side:

$$272n - 4n^2 + 2176 - 32n = 300n$$

$$-4n^2 + 240n + 2176 = 300n$$

$$-4n^2 - 60n + 2176 = 0$$

Divide throughout by $$-4$$:

$$n^2 + 15n - 544 = 0$$

Solve the quadratic:

$$n = \dfrac{-15 \pm \sqrt{225 + 2176}}{2} = \dfrac{-15 \pm \sqrt{2401}}{2} = \dfrac{-15 \pm 49}{2}$$

$$n = 17 \qquad \text{or} \qquad n = -32$$

Since the number of days cannot be negative, $$n = 17$$.

The work was actually completed in $$n + 8 = 17 + 8 = 25$$ days.