# Geometric Series Worksheet

Geometric Series Worksheet
• Page 1
1.
If $a$, $b$ and $c$ are in G.P as well as in A.P then
 a. $a$ = $b$ ≠ $c$ b. $a$ ≠ $b$ ≠ $c$ c. $a$ ≠ $b$ = $c$ d. $a$ = $b$ = $c$

#### Solution:

Given a, b, c are in G.P. So, b2 = ac ---------- (1)

Also given a, b, c are in A.P. So, b = a+c2 ---------- (2)

b2 = (a+c)24
[Square the equation (2) on both sides.]

(a+c)24 = ac
[Equate the values of b2, Use steps 1 and 3.]

(a + c)2 = 4ac
(a + c)2 - 4ac = 0
[Simplify.]

a2 + c2 + 2ac - 4ac = 0
(a - c)2 = 0
[a2 + c2 - 2ac = (a - c)2.]

(a - c) = 0, a = c
[Simplify.]

b = a+c2 =c+c2 = c
[Substitute the value of a in equation (2) to find b.]

Therefore, a = b = c
[From steps 7 & 8.]

2.
The value of ${14}^{\frac{1}{3}}\cdot {14}^{\frac{1}{9}}\cdot {14}^{\frac{1}{27}}\cdot$ ...................... ∞
 a. $\sqrt{14}$ b. 1 c. $\frac{1}{14}$ d. 14

#### Solution:

1413141914127 ...................... ∞
[As per the question.]

= 1413+19+127......................
[Use the formula: aman =am+n.]

= 1413(1-13)
[Use the formula: sum of infinite terms in geometric series = a1 - r.]

= 141323

= 1412

= 14

3.
If the common ratio of a GP is - $\frac{8}{25}$ and the sum to infinity is $\frac{1625}{33}$, then find the first term.
 a. 99 b. 65 c. 98 d. 67

#### Solution:

a1-r = 1625 / 33
[Formula for sum of infinite geometric series is a1-r where a is the first term and r is the common ratio.]

a1+825 = 1625 / 33
[Substitute r = - 8 / 25.]

a3325 = 1625 / 33
[Simplify.]

a = 1625 / 33× (33 / 25)
[Simplify.]

a = 65

Therefore, the first term = 65.

4.
If ${a}^{\frac{1}{x}}={b}^{\frac{1}{y}}={c}^{\frac{1}{z}}$ and $a$, $b$, $c$ are in G.P then $x$, $y$, $z$ are in ______
 a. None of these b. A.G.P c. G.P d. A.P

#### Solution:

Let a1x=b1y=c1z = k.

a = kx, b = ky, c = kz
[If a1x = k then a = kx.]

Given that a, b, c are in G.P so, b2 = ac.

k2y = kx. kz
[Substitute in step 3.]

k2y = kx + z
[When bases are equal powers should be added.]

2y = x + z
[Since bases on either sides are same, exponents should be equated.]

Therefore x, y, z are in A.P.

5.
What is the sum of the first 6 terms of the geometric series 1 + 4 + 16 + ... ?
 a. 16 b. 1365 c. 4 d. 1024

6.
Find the sum of 10 terms, S10 in the geometric series:
1 + $\frac{1}{3}$ + $\frac{1}{9}$ + $\frac{1}{27}$ + ...
 a. $\frac{3}{2}$(1 - $\frac{1}{{3}^{10}}$) b. $\frac{1}{3}$(1 - $\frac{1}{{3}^{10}}$) c. $\frac{2}{3}$(1 - $\frac{1}{{3}^{10}}$) d. $\frac{1}{{3}^{9}}$

7.
Find the sum of the infinite geometric series:
1 + $\frac{1}{9}$ + $\frac{1}{81}$ + ......
 a. $\frac{1}{9}$ b. $\frac{9}{8}$ c. $\frac{9}{10}$ d. $\frac{8}{9}$

8.
How many terms of the geometric sequence 7, 72, 73... are required to obtain a sum of 2800?
 a. 4 b. 3 c. 6 d. 5

9.
Find the common ratio of the geometric series, whose sum of first 39 terms, S39 = .
 a. 9 b. 39 c. 1 d. 10

 a. $\frac{1}{10}$ b. $\frac{11}{10}$ c. $\frac{10}{11}$ d. $\frac{1}{11}$