# 36. Geometric Progressions Solutions Part 1

1. Let $$a$$ be the first term and $$r$$ be the common ratio of the G. P. Let there are $$n$$ terms in the G. P. Then

$t_n = ar^{n - 1}$

It is evident that $$a = 5$$ and $$r = \frac{2 \text{nd term}}{1 \text{st term}} = \frac{3 \text{rd term}}{2 \text{nd term}} = 4$$

Thus, $$5.4^{n - 1} = 5120 \Rightarrow 4^{n - 1} = 1024 \Rightarrow n - 1 = 7 \Rightarrow n = 8$$.

So, we can conclude that there are 8 terms in this geometric progression.

2. Let $$a$$ be the first term and $$r$$ be the common ratio of the G. P.

$$\therefore ar^4 = 81$$ and $$ar = 24$$

Dividing, we have

$r^3 = \frac{81}{24} = \frac{27}{8} = -\frac{3}{2}$

Thus, $$ar = 24 \Rightarrow a = -16$$

It is evident that our geometric progression is $$-16, 24, 36, -54, 81, ...$$

3. This problem is similar to 1 and has been left as an exercise to the reader.

4. We have, per year increase as $$2\% = .02$$, thus, the common ratio becomes $$1.02$$ by which population will get multiplied every year. We have a period of 10 years. So from the formula for $$n~\text{th}$$ term, we have population in $$2016$$ as

$\Rightarrow 20000\times 1.02^{10 - 1} = 23902$
5. It is evident that $$a = 1$$ and $$r = 2$$. Thus, the person gets paid on $$20\text{th}$$ day, which, will be equal to $$20\text{th}$$ term of a G. P.

Thus, amount paid to him

$1\times 2^{19} = \524288$
6. Given $$S_n = 2^n$$ $$\therefore t_n = S_n - S_{n - 1} = 2^n - 1 - (2^{n - 1} - 1) = 2^{n - 1}$$

Let $$a$$ be the first term and $$r$$ be the common ratio of the geometric progression in question.

$$\therefore t_n = ar^{n - 1} = 2^{n - 1}$$ and $$t_{n - 1} = ar^{n -2} = 2^{n - 2}$$

Dividing $$t_n$$ by $$t_{n - 1}$$, we get common ratio

$r = \frac{t_n}{t_{n - 1}} = \frac{2^{n - 1}}{2^{n - 2}} = 2$

Substituting the value of $$r$$ in the formula for $$t_n$$, we get $$a.2^{n - 1} = 2^{n - 1}$$ i.e. $$a = 1$$.

Consequently, this series is in G. P.

7. Let $$a$$ be the first term and $$r$$ be the common ratio of the G. P.

$$\therefore t_7 = 8\times t_4$$ $$\Rightarrow ar^6 = 8ar^3 \Rightarrow r^3 = 8 \Rightarrow r = 2$$

Also, $$t_5 = 48 \Rightarrow ar^4 = 48 \Rightarrow a = 3 \because r = 2$$

Consequently, our G. P. is $$3, 6, 12, 24, 48, ...$$

8. Let $$a$$ be the first term and $$r$$ be the common ratio of the G. P. Also, let the three numbers are $$\frac{a}{r}, a, ar$$. Thus, product of these numbers is $$a^3 = 216 \Rightarrow a = 6$$. Now given that sum is $$19$$.

Thus,

$\frac{a}{r} + a + ar = 19 \Rightarrow \frac{6}{r} + 6 + 6r = 19$
$\Rightarrow 6r^2 + 6r + 6 = 19r \Rightarrow 6r^2 - 13r + 6 = 0$
$6r^2 - 9r - 4r + 6 = 0 \Rightarrow r = \frac{3}{2}, \frac{2}{3}$

When $$r = \frac{3}{2}$$ then the three numbers are $$4, 6, 9$$ and when $$r = \frac{2}{3}$$ then the three numbers are $$9, 6, 4$$.

9. Let the three digits be $$a, ar$$ and $$ar^2$$.

Given, $$a + ar^2 = 2ar + 1$$ or $$a(r - 1)^2 = 1$$.

Also given, $$a + ar = \frac{2}{3}(ar + ar^2)$$ or $$3a(1 + r) = 2ar(1 + r)$$

$$\Rightarrow (1 + r)(3 - 2r) = 0 \therefore r = \frac{3}{2}, -1$$

When $$r = \frac{3}{2}$$

$a = \frac{1}{(1 - r)^2} = \frac{1}{\left(\frac{3}{2} - 1\right)^2} = 4$

Following similarly, when $$r = -1 \Rightarrow a = \frac{1}{4}$$, which is not possible since $$a$$ is an integer.

Hence, $$a = 4, ar = 6$$ and $$ar^2 = 9$$.

$$\therefore$$ required number is 469.

10. Let the last three numbers in A. P. be $$b, b + 6, b + 12$$ and the first number be $$a$$.

Hence, the four numbers are $$a, b, b + 6$$ and $$b + 12$$.

From given conditions in questions, $$a = b + 12$$ and $$a, b, b + 6$$ are in G. P. i.e. $$b^2 = a(b + 6)$$

$$\Rightarrow b^2 = (b + 12)(b + 6)$$ $$\therefore b = -4$$

Thus, it follows that the four numbers are $$8, -4, 2$$ and $$8$$.

11. Let the three numbers are $$a, ar$$ and $$ar^2$$.

From given conditions, $$a + ar + ar^2 = 21$$ and $$a^2 + a^2r^2 + a^2r^4 = 189$$

From above equations we get,

$a(1 + r + r^2) = 21 ~~~~~~~ (i)$
$a(1 + r^2 + r^4) = 189 ~~~ (ii)$

Squaring (i) and dividing by (ii) by it, we get

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

or $$7(1 + r^2 -r) = 3(1 + r + r^2)$$ or $$4r^2 - 10r + 4 = 0$$

$$\therefore r = 2, \frac{1}{2}$$

Now it follows us that the numbers are $$3, 6, 12$$ or $$12, 6, 3$$.

12. Let $$a$$ be the first term and $$r$$ be the common ratio of the G. P., then

$t_6 = ar^5 = \frac{1}{16} ~~~~~~~~ t_{10} = ar^9 = \frac{1}{256}$

Dividing $$10\text{th}$$ term by $$6\text{th}$$ term, we get

$r^4 = \frac{1}{16} \Rightarrow r = \pm \frac{1}{2}$

When $$r = \frac{1}{2}$$, we have $$a = 2$$ and when $$r = -\frac{1}{2}$$, we have $$a = -2$$.

Thus, we have series as $$2, 1, x\frac{1}{2}, \frac{1}{4}, ...$$ or $$-2, 1, -\frac{1}{2}, \frac{1}{4}, ...$$

13. Let $$a$$ be the first term and $$r$$ be the common ratio of the G. P., then

$t_4 = ar^4 = 48 ~~~~~~~ t_7 = ar^6 = 384$

Dividing the $$8\text{th}$$ term by $$5\text{th}$$ term, we get

$$r^3 = 8 \Rightarrow r = 2$$. Substituting this in the equation for $$5\text{th}$$ term, we get

$$a.2^4 = 48 \Rightarrow a = 3$$. Thus, the geometric progression is $$3, 6, 12, 24, 48, ...$$

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

Given that first term is four times the third $$\therefore a = 4ar^2 \Rightarrow \pm \frac{1}{2}$$.

Also, given that product of three terms is $$-64$$ i.e. $$a . ar . ar^2 = -64 \Rightarrow ^3r^3 = -64$$. When $$r = \frac{1}{2}$$, $$a = -8$$ and when $$r = -\frac{1}{2}$$, $$a = 8$$.

$$\therefore$$ the G. P. is $$8, -4, 2, ...$$ or $$-8, -4, -2, ...$$

15. Here given sequence is a G. P. and $$a = 1, r = \frac{1}{2}$$.

$\therefore S_n = \frac{a(1 - r^n)}{1 - r} = 2\left[1 - \left(\frac{1}{2^n}\right)\right]$
16. Given series is $$a + a^2 + a^3 + ... + b(1 + 2 + 3 + ...)$$

$= \frac{a(1- a^n)}{1 - a} + \frac{bn(n + 1)}{2}$
17. Let $$S_n = 7 + 77 + 777 + ...~\text{to n terms}$$

$$= 7[1 + 11 + 111 + ...~\text{to n terms}]$$

$$= \frac{7}{9}[9 + 99 + 999 + ...~\text{to n terms}]$$

$$= \frac{7}{9}[(10 + 100 + 1000 + ...~\text{to n terms}) - (1 + 1 + 1 + ...~\text{to n terms})]$$

$=\frac{7}{9}\left[10\left(\frac{1 - 10^n}{1 - 10}\right) -n \right]$

$$= \frac{7}{81}(10^{n + 1} - 10 - 9n)$$

18. In the first case total salary $$= 1000 + 1000 + ...~\text{upto 31 terms} = \31000$$

In the second case, total salary $$= 1 + 2 + 4 + ...~\text{upto 31 terms} = \(2^{31} - 1)$$

Obviously, second case must be chosen.

19. Let the sum of $$n$$ terms of the given series be $$3280$$.

$\because S_n = \frac{a(1 - r^n)}{1 - r} \therefore 3280 = \frac{(3^n - 1)}{3 - 1}$

$$\Rightarrow 3^n = 6561 = 3^8 \Rightarrow n = 8$$

20. Given, $$1 + 3 + 3^2 + ... + 3^{n - 1} > 1000$$

$1.\left(\frac{3^n - 1}{3 - 1}\right) > 1000$
$3^n > 2000$

$$\therefore$$ the least value of $$n$$ for which $$3^n > 2001$$ is $$7$$.

21. $$.4\dot{2}\dot{3} = .423232323 ...~\text{to}~\infty$$

$= .4 + .23 + .0023 + .... ~\text{to}~\infty$
$= \frac{4}{10} + \frac{23}{1000}\left[1 + \frac{1}{100} + \frac{1}{10000} + ...\right]$
$= \frac{4}{10} + \frac{23}{1000}\left(\frac{1}{1 - \frac{1}{100}}\right)$
$= \frac{419}{990}$
22. Distance covered before the first reboud $$= 120$$ meters.

After striking the floor the ball rebounds to a height $$120.\frac{4}{5}$$ and then again falls back for the same height.

$$\therefore$$ distance covered between first and second reboud $$= 2 . 120 . \frac{4}{5}$$

Similarly, distance covered between second and third rebound $$= 2 . 120 . \left(\frac{4}{5}\right)^2$$.

$$\therefore$$ total distance covered before coming to rest

$= 120 + 2 . 120 . \frac{4}{5} + 2 . 120 . \left(\frac{4}{5}\right)^2 + ...~\text{to}~\infty$
$= 120 + 240\frac{4}{5}\frac{1}{1 - \frac{4}{5}}$

$$\therefore$$ total distance covered is 1080 meters.

23. $$S = 1 - 3 + 9 - 27 + ...~\text{to 9 terms}$$

It is evident that $$a = 1$$ and $$r = -3$$

$S = \frac{1 - r^n}{1 - r} = \frac{1 - (-3)^8}{1 - (-3)} = 4921$
24. It is evident that $$a = 1$$ and $$r = \frac{1}{3}$$

$S = \frac{1 - r^n}{1 - r} = \frac{3}{2}\left(1 - \frac{1}{3^n}\right)$
25. This problem is similar to 17 and has been left as an exercise.

26. This problem is similar to 17 and has been left as an exercise.

27. This problem is similar to 17 and has been left as an exercise.

28. We have to find sum of $$n$$ terms of the following series

$S = \left(x^2 + \frac{1}{x^2} + 2\right) + \left(x^4 + \frac{1}{x^4} + 5\right) + \left(x^6 + \frac{1}{x^6 + 8}\right) + ...$
$= x^2 + x^4 + x^6 + ... + \frac{1}{x^2} + \frac{1}{x^4} + \frac{1}{x^6} + ... + 2 + 5 + 8 + ...$
$= \frac{x^2(1 - x^{2n})}{1 - x^2} + \frac{1}{x^2}\frac{1 - \frac{1}{x^{2n}}}{1 - \frac{1}{x^2}} + \frac{n}{2}\left[4 + (n - 1)3\right]$
$= \frac{n(3n + 1)}{2} + \frac{x^{2n} - 1}{x^2 - 1}\left[x^2 + \frac{1}{x^{2n}}\right]$
29. Gievn, $$(p+q)$$ th term is $$a$$ and $$(p - q)$$ th term is $$b$$. Thus,

$t_{p + q} = t^{p + q -1} = a ~~~~~ t_{p - q} = t^{p - q -1} = b$

Multiplying both the terrms, we get

$t^{2(p - 1)} = ab \Rightarrow t^{p - 1} = \sqrt{ab}$

Thus, $$p$$ th term is $$\sqrt{ab}$$.

30. Let $$a$$ be the first term and $$c$$ be the common ratio of the G. P. Given

$x^{q - r}y^{r - p}z^{p - q} = 1$
$\Rightarrow (ac^{p - 1})^{(q - r)}(ac^{q - 1})^{(r - p)}(ac^{r - 1})^{(p - q)}$
$\Rightarrow a^{(q - r + r - p + p - 1)}c^{(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)}$
$\Rightarrow a^0c^0 = 1$
31. Let the three numbers $$a - d, a, a + d$$ are in A. P. Given that sum of these three terms is 15 $$\Rightarrow 3a = 15 \Rightarrow a = 5$$

Now after adding $$1 , 4$$ and $$19$$ the resulting numbers form a G. P. Thus, we have

$$(a - d + 1))(a + d + 19) = (a + 4)^2$$

Substituting the value of $$a$$, we get

$$\Rightarrow d^2 + 18d - 63 = 0$$

$$d = 3, -21$$

Thus numbers are $$2, 5, 8$$ or $$26, 5, -16$$

32. Let $$a$$ and $$b$$ be the first term and common ratio of the first G. P. respectively. Also, let $$x$$ and $$y$$ be the first term and common ratio of the second G. P. respectively.

Then, consider three terms from first A. P. $$\frac{a}{b}, a, ab$$ and three terms from second A. P. $$\frac{x}{y}, x, xy$$.

Given, that when these numbers are subtracted the resulting numbers form a G. P. as well. Thus, we have

$(a - x)^2 = \left(\frac{a}{b} - \frac{x}{y}\right)(ab - xy)$
$\frac{(a - x)^2}{ab - xy} = \frac{ay - bx}{by}$
$\frac{a^2 + x^2 - 2ax}{ab - xy} = \frac{ay - bx}{by}$
$a^2xy + x^2by - 2abxy = a^2by - ab^2x - axy^2 + x^by$
$ab^2x + axy^2 - 2axby = 0$
$ax(b - y)^2 = 0$

$$\Rightarrow b = y$$

Thus, we have proved that common ratios of two geormetric progressions are same. Now it remains to be proven that it is equal to the thirs sequence.

Common ratio of third seuqnece is:

$\frac{a - x}{\left(\frac{a}{b} - \frac{x}{y}\right)} = \frac{aby - bxy}{ay - bx}$

Substituting $$y = b$$

$\Rightarrow \frac{ab^2 - xb^2}{ab - bx} = b$

Thus, common ratios of all three sequences are same.

33. Let $$x$$ be the first term and $$y$$ be the common ratio of the G. P. then

$$a = xy^{p - 1}$$ $$\therefore \log a = \log x + (p - 1)\log y$$

$$b = xy^{q - 1}$$ $$\therefore \log b = \log x + (q - 1)\log y$$

$$c = xy^{r - 1}$$ $$\therefore \log c = \log x + (r - 1)\log y$$

Substituting in the equality for $$\log a, \log b$$ and $$\log c$$ we get the desired result.

34. Let $$r$$ be the common ratio of the given G. P. Since $$a, b, c, d$$ are in G. P.

$$\therefore b = ar, c = ar^2$$ and $$d = ar^3$$

$L. H. S. = (b - c)^2 + (c - a)^2 + (d - b)^2$
$= (ar - ar^2)^2 + (ar^2 - a)^2 + (ar^3 - ar)^2$
$= a^2[r^2 + r^4 - 2r^3 + r^4 - 2r^2 + 1 + r^6 - 2r^4 + r^2]$
$= a^2[r^6 - 2r^3 + 1] = (ar^3 - a)^2 = (d - a)^2 = (a - d)^2$
35. Let $$r$$ be the common ratio of the given G. P. Since $$a, b, c, d$$ are in G. P.

$$\therefore b = ar, c = ar^2$$ and $$d = ar^3$$

$L. H. S. = (a^2 + b^2 + c^2)(b^2 + c^2 + d^2)$
$= (a^2 + a^2r^2 + a^2r^4)(a^2r^2 + a^2r^4 + a^2r^6)$
$= a^2(1 + r^2 + r^4)a^2r^2(1 + r^2 + r^4)$
$= [a^2r(1 + r^2 + r^4)]^2$
$= [a.ar + ar. ar^2 + ar^2.ar^3]^2$
$= (ab + bc + cd)^2$
36. Given, $$a^x = b^y = c^z = k$$ (say).

Taling lagorithm, we get $$x\log a = y\log b = z\log c = \log k$$

$\therefore a = \frac{\log k}{\log a}, ~~ y = \frac{\log k}{\log b}, ~~ z = \frac{\log k}{\log c}$

Since $$x, y, z$$ are in G. P. $$\therefore \frac{y}{x} = \frac{z}{y}$$

$\therefore \frac{\log k}{\log b}\frac{\log a}{\log k} = \frac{\log k}{\log c}\frac{\log b}{\log k}$
$\Rightarrow \frac{\log a}{\log b} = \frac{\log b}{\log c}$

$$\Rightarrow \log_b a = \log_c b$$

37. Let $$a$$ be the first term and $$r$$ be the common ratio of the G. P. Then, we have

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

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

It is evident that $$(a + b)^2, (b + c)^2, (c + d)^2$$ are in G. P. having a common ratio of $$r^2$$.

38. Let $$a$$ be the first term and $$r$$ be the common ratio of the G. P. Then, we have

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

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

It is evident that $$(a - b)^2, (b - c)^2, (c - d)^2$$ are in G. P. having a common ratio of $$r^2$$.

39. Let $$a$$ be the first term and $$r$$ be the common ratio of the G. P. Then, we have

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

$a^2 + b^2 + c^2 = a^2(1 + r^2 + r^4)$
$ab + bc + cd = a^2r(1 + r^2 + r^4)$
$b^2 + c^2 + d^2 = a^2r^2(1 + r62 + r^4)$

It is evident that $$a^2 + b^2 + c^2, ab + bc + cd, b^2 + c^2 + d^2$$ are in G. P. having a common ratio of $$r$$.

40. These are just reciprocals of 37 and has been left as an exercise.

41. Let $$a$$ be the first term and $$r$$ be the common ratio of the G. P. Then, we have

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

$L. H. S. = a(b - c)^3 = a(ar - ar^2)^3 = a^4r^3(1 - r)^3$
$R. H. S. = d(a - b)^3 = ar^3(a - ar)^3 = a^4r^3(1 - r)^3$

It is evident that L. H. S. = R. H. S.

42. Let $$a$$ be the first term and $$r$$ be the common ratio of the G. P. Then, we have

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

$L. H. S. = (a + b + c + d)^2 = (a + ar + ar^2 + ar^3)^2$
$= a(1 + 2r + 3r^2 + 4r^3 + 3r^4 + 2r^5 + r^6)$
$R. H. S. = (a + b)^2 + (c + d)^2 + 2(b + c)^2$
$= a(1 + r^2 + 2r) + a(r^4 + 2r^5 + r^6) + a(2r^2 + 2r^4 + 4r^3)$
$= a(1 + 2r + 3r^2 + 4r^3 + 3r^4 + 2r^5 + r^6)$

It is evident that L. H. S. = R. H. S.

43. Let $$a$$ be the first term and $$r$$ be the common ratio of the G. P. Then, we have

Taking L. H. S. of the equality to be proven

$L. H. S. = a^2b^2c^2\left(\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3}\right)$
$= \frac{b^2c^2}{a} + \frac{a^2c^2}{b} + \frac{a^2b^2}{c}$
$= a^3r^6 + a^3r^3 + a^3 = c^3 + b^3 + a^3$
44. Let $$a$$ be the first term and $$r$$ be the common ratio of the G. P. Then, we have

Taking L. H. S. of the equality to be proven

$L. H. S. = (a^2 - b^2)(b^2 + c^2) = a^2(1 - r^2)a^2r^2(1 + r^2)$
$= a^4r^2(1 - r^2)(1 + r^2) = (a^2r^2 - a^2r^4)(a^2 + a^2r^2)$
$= (b^2 - c^2)(a^2 + b^2) = R. H. S.$
45. Let $$a$$ be the first term and $$r$$ be the common ratio of the G. P. Then, we have

$$S = a + ar + ar^2 + ... \text{to n terms}$$

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

Again, $$P = a . ar . ar^2 ... \text{to n terms}$$

$$= a^n . r^{1 + 2 + ... + n} = a^nr^{\frac{n(n - 1)}{2}}$$

Also,

$R = \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + ... \text{to n terms}$
$= \frac{\frac{1}{a}\left[1 - \left(\frac{1}{r}\right)^n\right]}{1 - \frac{1}{r}}$
$= \frac{1}{a}.\frac{r^n - 1}{r^n}.\frac{r}{r - 1}$
$R = \frac{1 - r^n}{1 - r}.\frac{1}{ar^{n - 1}}$
$\frac{S}{R} = a^2r^{n - 1}$
$\left(\frac{S}{R}\right)^n = (a^2r^{n - 1})^n = a^{2n}r^{n(n - 1)} = P^2$
46. The first term $$a = 1$$ and the common ratio $$r = \frac{1}{2}.$$

$\therefore S_{\infty} = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{2}} = 2$
47. What we have to prove can be written as

$\frac{t_n}{S_{\infty} - S_n} = \frac{1 - r}{r}$
$L. H. S. = \frac{ar^{n - 1}}{\frac{a}{1 - r} - \frac{a(1- r^n)}{1 - r}}$
$= \frac{ar^{n - 1}}{\frac{a}{1 - r}(1 -1 + r^n)} = \frac{ar^{n - 1}}{\left(\frac{a}{1 - r}\right)r^n}$
$= \frac{1 - r}{r}$
48. Sum of inifinite terms of a G. P. whose common ratio is numerically less than $$1$$ is given by

$$S_{\infty} = \frac{a}{1 - r}$$ where $$a$$ is the first term and $$r$$ is the common ratio. Therefore,

$S_1 = \frac{1}{1 - \frac{1}{2}} = 2 ~~~ \left[\because a = 1, r = \frac{1}{2}\right]$
$S_2 = \frac{2}{1 - \frac{1}{3}} = 3 ~~~ \left[\because a = 2, r = \frac{1}{3}\right]$
$...$
$S_p = \frac{p + 1}{1 - \frac{1}{p}} = p + 1 ~~~ \left[\because a = p, r = \frac{1}{p}\right]$

Now, $$L. H. S. = S_1 + S_2 + ... + S_p$$

$$= 2 + 3 + ... + (p + 1) = \frac{p}{2}[2.2 + (p - 1).1] = \frac{p(p +3)}{2}$$

49. $$\because S_{\infty} = \frac{a}{1 - r}$$ if $$|r| < 1$$, here common ratio is $$a$$ and $$0 < a < 1$$

$\therefore x = 1 + a + a^2 + ... \text{to}~\infty = \frac{1}{1 - a}$
$\therefore a = \frac{x - 1}{x}~~~~~~~~(i)$

Similarly,

$y = 1 + b + b^2 + ... \text{to}~\infty = \frac{1}{1 - b}$
$\therefore b = \frac{y - 1}{y}~~~~~~~(ii)$

Now since, $$0 < a < 1$$ and $$0 < b < 1$$ therefore $$0 < ab < 1$$

Now,

$1 + ab + a^2b^2 + ... \text{to}~\infty = \frac{1}{1 - ab}$

Substititing the values of $$a$$ and $$b$$ from (i) and (ii), we get

$= \frac{1}{1 - \frac{x - 1}{x}.\frac{y - 1}{y}} = \frac{xy}{x + y + 1}$
50. $$1 + (1 + a)r + (1 + a + a^2)r^2 + ... \text{to}~\infty$$

$= \frac{1 - a}{1 - a} + \frac{1 - a^2}{1 - a}r + \frac{1 - a^3}{1 - a}r^2 + ... \text{to}~\infty$
$= \frac{1}{1 - a}[(1 - a) + (1 - a^2)r + (1 - a^3)r^2 + ... \text{to}~\infty]$
$= \frac{1}{1 - a}[(1 + r + r^2 + ... \text{to}~\infty) - a(1 + a + (ar)^2 + ... \text{to}~\infty)]$
$= \frac{1}{1 - a}\left[\frac{1}{1 - r} - a \left(\frac{1}{1 - ar}\right)\right]$
$= \frac{1}{(1 - r)(1 - ar)}$