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 + ar^{2} + ... + ar^{n–2} + ar^{n–1} ... (1)
Multiplying (1) byr, we get
r S_{n} = ar + ar^{2} + .... + ar^{n–2} + ar^{n–1} + ar^{n} ... (2)
(1) – (2) ⇒ S_{n }– rS_{n} = a – ar^{n}
or S_{n} (1 – r) = a (1 – r^{n})
.....(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)^{n2} = 3^{4} = (√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)
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