The document Sum of n-terms of a G.P. CA Foundation Notes | EduRev is a part of the CA Foundation Course Quantitative Aptitude for CA CPT.

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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^{nâ€“2} + ar^{nâ€“1} ... (1)

Multiplying (1) byr, we get

r S_{n} = ar + ar^{2} + .... + ar^{nâ€“2} + ar^{nâ€“1} + ar^{n} ... (2)

(1) â€“ (2) â‡’ S_{n }â€“ rS_{n} = a â€“ ar^{n}

or S_{n} (1 â€“ r) = a (1 â€“ r^{n})

.....(A)

.......(B)

Either (A) or (B) gives the sum up to the n^{th }term when r â‰ 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/âˆš3, 1, âˆš3, .......,81**

**Solution **: Here, a =** **1/âˆš3 , r = âˆš3 and t_{n} = l = 81

âˆ´ (âˆš3)^{n-2} = 3^{4} = (âˆš3)^{8}

âˆ´ n â€“ 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 [âˆµ 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{(1â€“0.1) + (1â€“0.01) + (1 â€“ 0.001) + to n terms}

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

...... (Since r < 1)

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