Sum of n-terms of a G.P. | Quantitative Aptitude for CA Foundation PDF Download

Let a denote the first term and r the common ratio of  a G. P. Let S_n represent the sum of first n terms of the G. P.

Thus, S_n = a + ar + ar² + ... + ar^n–2 + ar^n–1 ...   (1)

Multiplying (1) byr, we get
 r S_n = ar + ar² + .... + ar^n–2 + ar^n–1 + arⁿ ... (2)

(1) – (2) ⇒ S_n – rS_n = a – arⁿ

or S_n (1 – r) = a (1 – rⁿ)

 .....(A)

  .......(B)

Either (A) or (B) gives the sum up to the n^th term when r ≠ 1. It is convenient to use formula (A) when | r | < 1 and (B) when | r | >1.

Example 1. Find the sum of the G. P.: 1, 3, 9, 27, ... up to the 10^th term.

Solution : Here the first term (a) = 1 and the common ratio (r) = 3/1 = 3

Now using the formula, 

Example 2. Find the sum of the G. P.: 1/√3, 1, √3, .......,81

Solution : Here, a = 1/√3 , r = √3 and t_n = l = 81

∴ (√3)^n-2 = 3⁴ = (√3)⁸

∴ n – 2 = 8

or n =10

Example 3. Find the sum of the G. P.:  0.6, 0.06, 0.006, 0.0006, ........ to n terms.

Solution : here, a = 0.6 = 6/10 and r = 0.06/0.6 = 1/10

Using the formula     we have [∵ r <1]

Hence, the required sum is 

Example 4. How many terms of the following G. P.: 64, 32, 16, ...... has the sum 

Solution : here, a = 64, r = 32/64 = 1/2 (< 1), and

Using the formula    , we get

.... .. (given)

n = 8 

Thus, the required number of terms is 8.

Example 5.  Find the sum of the following sequence :  2, 22, 222, ......... to n terms.

Solution :  Let S denote the sum. Then 

S = 2 + 22 + 222 + ..... to n terms

= 2 (1 + 11 + 111 + .... to n terms)

= 2/9 (9 + 99 + 999 + .... to n terms)

Example 6. Find the sum up to n terms of the sequence:  0.7, 0.77, 0.777, .......

Solution : Let S denote the sum, then

S = 0.7 + 0.77 + 0.777 + ...... to n terms 

= 7(0.1 + 0.11 + 0.111 + ...... to n terms)

= 7/9 (0.9 + 0.99 + 0.999 + ..... to n terms)

= 7/9{(1–0.1) + (1–0.01) + (1 – 0.001) +
to n terms}

= 7/9{(1 + 1 + 1 + ... n terms) – (0.1 + 0.01 + 0.001 +
to n terms)}

  ...... (Since r < 1)