# 31. Geometric Progressions Solutions Part 2¶

1. Let $$r$$ be the common ratio then $$b = ar^{n - 1}$$. And we have $$p$$ as

$p = a.ar.ar^2. ... .ar^{n - 1}$
$p = a^nr^{\frac{n(n - 1)}{2}}$
$p^2 = a^{2n}r^{n(n - 1)}$
$p^2 = (a^2r^{n - 1})^n$
$p^2 = (a.ar^{n - 1})^n$
$p^2 = (ab)^n$
2. Let $$a_1$$ and $$a_2$$ be the first terms of the two G. P.s are $$r$$ be the common ratio. Let $$S_1$$ and $$S_2$$ be the sum to $$n$$ terms, we have

$S_1 = \frac{a_1(1 - r^{n - 1})}{1 - r}~~~S_2 = \frac{a_2(1 - r^{n -1})}{1 -r}$
$\therefore \frac{S_1}{S_2} = \frac{a_1}{a_2} = \frac{a_1r^{n - 1}}{a_2r^{n - 1}}$

Thus, ratio of sums to $$n$$ terms is equal to the $$n\text{th}$$ terms.

3. Let $$a$$ be the first term and $$r$$ be the common ratio. Then, we have

$S_1 = a + ar + ar^2 + ... + ar^{n - 1}$
$S_2 = a + ar + ar^2 + ... + ar^{n - 1} + ar^n + ar^{n + 1} + ... + ar^{2n - 1}$
$S_3 = a + ar + ar^2 + ... + ar^{2n - 1} + ar^{2n} + ... + ar^{3n - 1}$
$S_2 - S_1 = ar^n + ar^{n + 1} + ... + ar^{2n - 1}$
$S_2 - S_1 = \frac{ar^n(r^{n - 1} - 1)}{r - 1}$

Similarly,

$S_3 - S_2 = \frac{ar^{2n}(r^{n - 1} - 1)}{r - 1}$

and

$S_1 = \frac{a(r^{n - 1} - 1)}{r - 1}$

Thus, $$(S_2 - S_1)^2 = S_1(S_3 - S_2)$$.

4. We need to compute following:

$S_1 + S_2 + ... + S_{2n - 1}$

From formula for sum of geometric series we have following:

$= \frac{a(r - 1)}{r - 1} + \frac{a(r^2 - 1)}{r - 1} + ... + \frac{a(r^{2n - 1} - 1)}{r - 1}$
$= \frac{a}{r-1}\left[r + r^2 + ... + r^{2n - 1} - (2n - 1)\right]$
$= \frac{a}{r-1}\left[\frac{r(1 - r^{2n - 1})}{r - 1} - (2n - 1)\right]$
5. Given, $$S_n = a.2^n - b$$ $$\therefore S_{n - 1} = a.2^{n - 1} - b$$. Thus, $$t_n = S_n - S_{n - 1} = a2^{n - 1}$$. Since the ratio of terms will be 2 as evident from $$t_n$$ the series is in G. P.

6. The sum would be $$S = 3.2 - 4 + 3.2^2 - 4 + ... + 3.2^{100} - 4$$.

From the formula for sum of a geometric series, we have,

$$S = 6(2^{100} - 1) - 400$$

7. $$t_n = 1 + 2 + 2^2 + ... + 2^n = 2^n - 1$$.

Sum of $$n$$ terms = $$S = 1 + 3 + 7 + 15 + ... + 2^n - 1$$

$S = 2^1 - 1 + 2^2 - 1 + 2^3 - 1 + 2^n - 1$
$S = 2^{n + 1} - 2 - n$
8. Clearly, the common ratio is $$3x$$. Thus, from the formula for sum of infinite series we have sum as

$S_{\infty} = \frac{1}{1 - 3x}$
9. Common ratio is $$\frac{-1}{3}$$, thus sum for infinite terms of the given series is:

$S_{\infty} = \frac{1}{1 - (\frac{1}{3})} = \frac{9}{4}$
10. There are two serieses having common ratios $$\frac{1}{5}$$ and $$\frac{1}{7}$$ having first terms as $$\frac{1}{5}$$ and $$\frac{1}{7}$$ respectively, as well.

Thus, applying the formula, we have

$S = \frac{\frac{1}{5}}{1 - \frac{1}{5}} + \frac{\frac{1}{7}}{1 - \frac{1}{7}}$
$S = \frac{1}{4} + \frac{1}{6} = \frac{5}{12}$
11. Value of exponent is: $$S = \frac{1}{3} + \frac{1}{9} + \frac{1}{23} + ... ~~\text{to}~~ \infty$$

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

Thus, value is $$9^\frac{1}{2} = 3$$.

12. Sum within braces is

$S = \frac{1}{1 - \frac{2x}{1 + x^2}} = \frac{1 + x^2}{(1 - x)^2}$

Thus, final sum is $$S = \frac{1}{(1 - x^2)}$$.

13. $$L. H. S. = a^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... ~~\text{to}~~ \infty}$$

The value of exponent is $$\frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1$$. Thus final value is $$a$$.

14. $$0.\dot{5}\dot{4} = 0.54 + 0.0054 + 0.000054 + ... ~~\text{to}~~ \infty$$

Thus, we have

$0.\dot{5}\dot{4} = \frac{54}{100} + \frac{54}{10000} + \frac{54}{1000000} + ...$
$= \frac{54}{100}.\frac{1}{1 - \frac{1}{100}} = \frac{54}{99} = \frac{6}{11}$
15. We have

$.\dot{6} = .6 + .06 + .006 + ... ~~\text{to}~~ \infty$
$= \frac{6}{10} + \frac{6}{100} + \frac{6}{1000} + ...$
$= \frac{6}{10}.\frac{1}{1 - 10} = \frac{6}{9} = \frac{2}{3}.$
16. Given, $$S_{\infty} = 32$$ and $$a + ar = 24$$, where, $$a$$ is the first term and $$r$$ is the common ratio.

Now, $$S_{\infty} = \frac{a}{1 - r} \because |r| < 1.$$

Solving these two we have,

$a = 32(1 - r) ~~\text{and}~~ a(1 + r) = 24$

Substituting first in second we have

$32(1 - r^2) = 24 ~~\text{i.e.}~~ r^2 = \frac{1}{4} \Rightarrow r = \pm\frac{1}{2}$

Thus, $$a = 48$$ or $$a = 16$$. Thus, we can have our desired series as $$16, 8, 4, 2, ...$$ or $$48, -24, 12, -6$$.

17. Let $$a$$ be the first term and $$r$$ be the common ratio. Then, we have

$\frac{a}{1 - r} = 4 ~~\text{and}~~ \frac{a^2}{1 - r^2} = \frac{16}{3}$

Substituting first ins second, we have

$\frac{16(1 - r)^2}{1 - r^2} = \frac{16}{3} \Rightarrow \frac{1 - r}{1 + r} = \frac{1}{3}$

Solving, we have $$r = \frac{1}{2}$$ and then $$a = 2$$. Thus, we have our series as $$2, 1, \frac{1}{2}, \frac{1}{4}, ...$$.

18. Let $$a$$ be the first term and $$r$$ be the common ratio. Then, we have

$S = 1 + a + ar + ar^2 + ... ~~\text{to}~~ \infty$

Let us consider a term $$t_n = ar^{n - 1}$$. Then, sum of succeeding terms is

$$S = \frac{ar^n}{1 - r}$$. Equating these we have

$\frac{ar^n}{1 - r} = ar^n - 1$
$\Rightarrow \frac{r}{1 - r} = 1 \Rightarrow r = \frac{1}{2}.$

Similarly we can prove for $$>$$ or $$<$$.

19. Given, $$y = 1 + x + x^2 + ...$$

From the formula for infinite G. P.

$y = \frac{1}{1 - x} \Rightarrow 1 - x = \frac{1}{y} \Rightarrow x = \frac{y - 1}{y}$
20. $$c = ar^2$$ and $$b = ar$$. Thus, we have $$c > 4b - 3a = ar^2 > 4ar - 3a$$

$$\Rightarrow r^2 > 4r - 3 \Rightarrow (r - 1)(r - 3) > 0$$

Thus, $$r > 3$$ or $$r < 1$$.

21. $$1 + 2x + 4x^2 + ... + 32x^5 = \frac{1 - k^6}{1 - k}$$

$$\frac{1 - (2x)^6}{1 - 2x} = \frac{1 - k^6}{1 - k}$$

$$\therefore k = 2x \Rightarrow \frac{k}{x} = 2.$$

22. Simplifying the given relation, we have

$(b^4 - 2b^2ac + a^2c^2) + (c^4 - 2c^2bd + b^2d^2) + (a^2d^2 - 2abcd - b^2c^2) \le 0$
$\Rightarrow (b^2 - ac)^2 + (c^2 - bd)^2 + (ad - bc)^2 \le 0$

This is only possible if and only if, $$b^2 = ac, c^2 = bd, ac = bd$$ i.e. $$\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$$. Thus, $$a, b, c, d$$ are in G. P.

23. On simplification, we have

$\sum (a_1^2a_3^2 + a_2^4 - 2a_2^2a_1a_3) \le 0$

or $$\sum (a_1a_3 - a_2^2)^2 \le 0$$.

In L. H. S. every bracket being square of real number is +ve and hence their sum cannot be less than zero. It will be zero if each term is zero.

$$\therefore a_1a_3 = a_2^2$$ or $$a_1, a_2, a_3$$ are in G. P. Similarly, others are also in G. P.

Hence proved.

24. $$\alpha, beta, \gamma, \delta$$ being in increasing G. P., they may be taken as $$k, kr, kr^2, kr^3$$, where $$r > 1$$.

Sum of the roots of the equations

$S_1 = k(1 + r) = 3, S_2 = kr^2(1 + r) = 12$

Substituting $$S_1$$ in $$S_2$$, we have

$$3r^2 = 12$$ or $$r = 2 \therefore k = 1$$.

Product of the roots

$$P_1 = \alpha\beta = k^2r = a, P_2 = \gamma\delta = k^2r^5 = b$$

Putting for $$k$$ and $$r$$, we have

$$a = 2, b = 32$$.

25. Given $$d = 2, r = \frac{1}{2}.$$

If there be odd number of terms then mid-term = $$\frac{1}{2}(odd + 1)$$.

$$T_{2n + 1}$$ is the mid-term of sequence of $$(4n + 1)$$ odd terms.

$$a + 2nd = a + 4n$$.

This middle term is the last term of A. P. and first term of following G. P. each of $$(2n + 1)$$ terms with this term being common to both.

Let $$T_{n+1}$$ and $$t_{n + 1}$$ are mid terms of A. P. and G. P.

$$T_{n + 1} = a + nd = a + 2n$$

$$t_{n + 1} = AR^n = T_{2n + 1}\left(\frac{1}{2}\right)^n = (a + 4n)\left(\frac{1}{2}\right)^n$$

Given, $$T_{n + 1} = t_{n + 1}$$

$$a + 2n = (a + 4n)\frac{1}{2^n}$$

$\therefore a = \frac{4n - n.2^{n + 1}}{2^n - 1}$

Hence, mid term is

$a + 4n = \frac{n.2^{n + 1}}{2^n - 1}$
26. $$S_n = 3 - \frac{3^{n + 1}}{4^{2n}}$$. Putting $$n = 1, 2$$ we have

$T_1 = S_1 = 3 - \frac{9}{16} = \frac{39}{16}$
$S_2 = 3 - \frac{27}{256} = T_1 + T_2$
$\therefore T_2 = \frac{117}{256}$
$r = \frac{T_2}{T_1} = \frac{3}{16}$
27. Let three number in G. P. are $$ar, a, \frac{a}{r}$$ then given that $$ar, 2a, \frac{a}{r}$$ in A. P.

$\therefore 2(2a) = a\left(r + \frac{1}{r}\right)$

or $$r^2 - 4r + 1 = 0$$

or $$r = 2 \pm \sqrt{3}$$

Since, it is an increasing G. P. therefore $$r = 2 + \sqrt{3}$$.

28. Given, $$(4x + 1)^2 = (2x + 1)(8x + 1)$$

$$2x = 0 \Rightarrow x = 0$$

But $$x = 0$$ makes $$f(x), f(2x)$$ and $$f(4x)$$ equal which is G. P. of $$r = 1$$.

29. Let $$r$$ be the common ratio, then, we have

$$a + ar + ar^2 = x. ar$$ or $$r^2 + r(1 - x) + 1 = 0$$, $$r$$ is real.

$$Discriminant > 0$$ i.e. $$(1 - x)^2 - 4 > 0$$

or $$(x + 1)(x - 3) > 0$$ $$\Rightarrow x < -1 ~~\text{or}~~ x > 3$$.

30. Let the numbers be $$a$$ and $$b$$ then $$A = \frac{a + b}{2}$$ or $$a + b = 2A$$

Also, $$G = \sqrt{ab} \Rightarrow G^2 = ab$$

Thus, $$a$$ and $$b$$ are roots of

$$t^2 - 2At + G^2 = 0$$

$t = \frac{2A \pm \sqrt{4A^2 - 4G^2}}{2} = A \pm \sqrt{A^2 - G^2}$
31. Let $$A$$ be the A. M. and $$G$$ be the G. M., then, we have

$\frac{A}{G} = \frac{m}{n}$
$\frac{A}{m} = \frac{G}{n} = k$

$$\therefore a + b = 2mk, ab = n^2k^2$$

Hence, $$a, b$$ are roots of

$$x^2 - 2mkx + k^2n^2 = 0$$

$$\therefore x = 2mk \pm 2k\sqrt{m^2 - n^2}$$

$$\therefore a:b = m + \sqrt{m^2 - n^2}: m - \sqrt{m^2 - n^2}$$.

32. $$x = \frac{1}{1 - a}, y = \frac{1}{1 - b}, z = \frac{1}{1 - c}$$

$$\therefore \frac{1}{x} = 1 - a, \frac{1}{y} = 1 - b, \frac{1}{z} = 1 - c$$

Since $$a, b, c$$ are in A. P., therefore $$\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$$ are in A. P.

33. For $$p, a =1, r = -\tan^2x$$

$\therefore p = \frac{a}{1 - r} = \frac{1}{1 + \tan^2x} = \cos^2x$

For $$q, a =1, r = -\cot^2y$$

$\therefore q = \frac{a}{1 - r} = \frac{1}{1 + \cot^2y} = \sin^2y$
$S = \frac{1}{1 - \tan^2x\cot^2y}$
$= \frac{1}{1 - \frac{1 - \cos^2x}{\cos^2x}\frac{1 - \sin^2y}{\sin^2y}}$
$= \frac{pq}{p + q + q} = \frac{1}{\frac{1}{p} + \frac{1}{q} - \frac{1}{pq}}$
34. Let side of outermost equilateral triangle is $$a$$, then its area is $$\frac{\sqrt{3}}{4}a^2$$. The sides of subsequent internal triangles will be $$\frac{a}{2}, \frac{a}{4}, \frac{a}{8}, ...$$

Therefore, total area is $$\frac{\sqrt{3}}{4}a^2\left(\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ... \right)$$

$= \frac{\sqrt{3}}{4}a^2. \frac{1}{1 - \frac{1}{4}} = \frac{\sqrt{3}}{\frac{a^2}{3}} = 1$
35. $$\cos^2x = |\cos^2x|$$

Sum of infinite series is $$S = \frac{1}{1 - |\cos x|}$$ where $$|\cos x| < 1$$.

$$E = e^{S\log_e 4} = 4^S$$

$$E$$ satisfied theq equation $$t^2 - 20t + 64 = 0 \therefore t = 16, 4$$

$$4^S = 4^1$$ or $$4^2$$ or $$S = 1$$ or $$2$$

or $$\frac{1}{1 - |\cos x|} = 1 ~~\text{or}~~ 2$$

or $$1 - |\cos x| = 1 ~~\text{or}~~\frac{1}{2}$$

$$\Rightarrow |\cos x| = 0 ~~\text{or}~~ \frac{1}{2}$$

$$\therefore \cos x = 0 ~~\text{or}~~ \pm\frac{1}{2}$$

$$x = \frac{\pi}{2}, \frac{\pi}{3}, \frac{2\pi}{3}$$.

36. The given equation may be written as

$$8^{1 + |\cos x| + |\cos^2x| + .. ~~\text{to}~~ \infty} = 8^2$$

$$1 + |\cos x| + |\cos^2x| + .. ~~\text{to}~~ \infty = 2$$

To sum the G. P., we must observe that for

$$-\pi < x < \pi, x \ne 0$$, we have $$|\cos x| < 1$$.

Hence $$\frac{1}{1 - |\cos x|} = 2$$ or $$1 - |\cos x| = \frac{1}{2}$$ by $$S_{\infty}$$ for G. P.

or $$|\cos x| = 1/2$$ i.e. $$\cos x = \pm 1/2$$.

37. $$T_n = (1 + a + a^2 + ... + a^{n - 1})b^{n - 1}$$

$= \frac{1 - a^n}{1 - a}.b^{n - 1}$
$= \frac{1}{1 - a}[b^{n - 1} - a(ab)^{n - 1}]$

Putting $$n = 1, 2, 3, ..., \infty$$ and adding

$S_{\infty} = \frac{1}{1 - a}[(1 + b + b^2 + ... \infty) + a(1 + ab + a^2b^2 + ... \infty)]$
$S_{\infty} = \frac{1}{1 - a}\left[\frac{1}{1 - b} - a.\frac{1}{1 - ab}\right]$
$S_{\infty} = \frac{1}{(1 - b)(1 - ba)}$
38. $$S_{\infty} = \frac{\sin^2x}{1 - \sin^2x} = \tan^2x$$

$L. H. S. = e^{\tan^2x \log 2} = 2^{\tan^2x}$

and the roots of the equation $$x^2 - 9x + 8 = 0$$ are $$1$$ and $$8$$.

$$2^{\tan^2x} = 1 = 2^0, 2^{\tan^2x} = 8 = 2^3$$

$$\therefore \tan^2x = 0, \tan^2x = 3$$

$$\therefore \tan x = 0, \tan x = \pm \sqrt{3}$$

$$\therefore x = \frac{\pi}{3}$$ is the only value of $$x$$ satisfying $$0 < x < \frac{\pi}{2}$$

$$\therefore \frac{\cos x}{\cos x + \sin x} = \frac{1}{1 + \tan x} = \frac{1}{1 + \sqrt{3}}$$

39. $$S_{\lambda = 1 + \frac{1}{\lambda} + \frac{1}{\lambda^2} + ... \infty} = \frac{\lambda}{\lambda - 1}$$

$\therefore \sum_{\lambda = 1}^n (\lambda - 1)S_{\lambda} = \sum_{\lambda}^n \lambda = \frac{n(n + 1)}{2}$
40. $$\frac{T_2}{T_1} = \frac{T_3}{T_2} \Rightarrow 2^{(b - a).x} = 2^{(c - b)x}$$

$$\Rightarrow (b - a)x = (c - a)x \Rightarrow b - a = c - a \forall x, x\ne 0$$

Above is true as $$a, b, c$$ are in A. P.

41. Writing, $$a + be^x = 2a - (a - be^x)$$, we have

$\frac{2a}{a - be^x} - 1 = \frac{2b}{b - ce^x} - 1 = \frac{2c}{c - de^x} - 1$
$\Rightarrow \frac{a - be^x}{a} = \frac{b - ce^x}{b} = \frac{c - de^x}{c}$
$\Rightarrow 1 - \frac{b}{a}e^x = 1 - \frac{c}{b}e^x = 1 - \frac{d}{c}e^x$
$\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$

Thus, $$a, b, c$$ are in G. P.

42. Since, $$x, y, z$$ are in G. P. $$y^2 = xz$$

and $$2\tan^{-1}x = \tan^{-1}y + \tan^{-1}z$$

$\frac{2y}{1 - y^2} = \frac{x + z}{1 - xz} \Rightarrow 2y = x + z$
$4y^2 = (x + z)^2 \Rightarrow 4zx = (x + z)^2 \Rightarrow (x - z)^2 = 0 \Rightarrow x = z$
$\therefore x = y = z$
43. Given, $$b - a = c - b$$ and $$(c - b)^2 = a(b - a)$$

or $$(b - a)^2 = a(b - a)^2 \Rightarrow b = 2a$$

but $$c = 2b - a = 3a$$.

$$\therefore a : b : c = 1 : 2 : 3$$

44. $$\log \frac{a}{2b}, \log \frac{2b}{3c}, \log \frac{3c}{a}$$ are in A. P.

$$\therefore 2\log \frac{2b}{3c} = \log \frac{a}{2b} + \log \frac{3c}{a}$$

$$\log\left(\frac{2b}{3c}\right)^2 = \log\left(\frac{a}{2b}.\frac{3c}{a}\right)$$

$$\Rightarrow \frac{4b^2}{9c^2} = \frac{3c}{2b}$$ or $$8b^3 = 27c^3$$ $$\therefore 2b = 3c$$

Also, $$a, b, c$$ are in G. P. $$\therefore b^2 = ac$$

$$\frac{9c^2}{4} = ac$$ $$\therefore a = \frac{9}{4}c$$

Thus, sides are $$\frac{9}{4}c, \frac{6}{4}c$$ and $$c$$. Clearly, $$a$$ is greatest side so that $$\angle A$$ is greatest.

$$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = -\frac{29}{48} < 0$$

Therefore, $$\angle A$$ is obstuse so the triangle is obtuse angled triangle.

45. Given,

$\begin{split}Area = \begin{vmatrix} a & c & e & a \\ d & e & f & b \end{vmatrix}\end{split}$

Substituting the values and evaluating the determinant will yield the desired result.

46. $$a^t = \log_t a . \log_b t = \log_b a$$

$$t = \log_a(\log_b a)$$