## GEOMETRIC SERIES WORKSHEET

Geometric series worksheet :

Here you are going to see practice questions of the topic geometric series. You can find answer for each questions in the page below.

## Geometric series worksheet - Practice questions

 Questions Solution (1) Find the sum of first 20 terms of the geometric series 5/2 + 5/6 + 5/18 + ............ Solution (2) Find the sum of first 27 terms of the geometric series 1/9 + 1/27 + 1/81 + ............ Solution (3) Find the sum of n terms of the geometric series described below (i) a = 3,t₈ = 384,n=8  (ii) a = 5,r = 3,n=12 Solution (4) Find the sum of the following finite series (i) 1 + 0.1 + 0.01 + 0.001 + .......... + (0.1)⁹(ii) 1 + 11 + 111 + .............. to  20 terms Solution (5) How many consecutive terms starting from the first term of the  series (i) 3 + 9 + 27 + ........ would be 1092?(ii) 2 + 6 + 18 + ........... would be 728? Solution (6) The second term of the geometric series is 3 and the common ratio is 4/5. Find the sum of first 23 consecutive terms in the given geometric series. Solution (7) A geometric series consists of four terms and has a positive common ratio. The sum of the first two terms is 9 and the sum of the last two terms is 36. Find the series. Solution (8) Find the sum of the first n terms of the geometric series(i) 7 + 77 + 777 + ..............(ii) 0.4 + 0.94 + 0.994 + ............ Solution (9) Suppose that five people are ill during the first week of an epidemic and each sick person spreads the contagious disease to four other people by the end of the second week and so on. By the end of 15th week,how many people will be affected by the epidemic? Solution (10) A gardener wanted to reward a boy for his good deeds by giving some mangoes. He gave the boy two choices. He could either have 1000 mangoes at once or he could get 1 mango on the first day,2 on the second day,4 on the third day,8 mangoes on the fourth day and so on for ten days. Which option should the boy choose to get the maximum number of mangoes? Solution (11) A geometric series consist of even number of terms. The sum of all terms is 3 times the sum of odd terms. Find the common ratio. Solution (12) If S₁,S₂ and S₃ are the sum of first n, 2n and 3n terms of a geometric series respectively then prove that S₁ (S₃ - S₂) = (S₂ - S₁)² Solution

(1)  (-15/4) [(1/3)20 - 1 ]

(2)  (1/6) [1 - (1/3)27]

(3)  (i) 765    (ii) S₁₂ = (5/2) (312 - 1)

(4)  (i) [1- (0.1)10]/0.9     (ii) 10/81(1020 - 1) - 20/9

(5)  (i) n = 6  (ii)  n = 6

(6)  (75/4) (1-(4/5)23)

(7)  3 + 6 + 12 + 24 + ......

(8)  (i) 70/81 (10n - 1) - 7n/9   (ii) n  - 2/3 [1 - (0.1)n]

(9)  S15 = (5/3) (415-1)

(10)  1023, second way

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Sequence and Series

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