# Find common ratio when sum of n terms of geometric progression is given

In this section you will learn to find common ratio when sum of n terms of geometric progression is given.Here we will use the following formulas :

1) $a_{n} = ar^{n - 1}$

2) $S_{n} = a_{1}\left ( \frac{r^{n} - 1}{r - 1} \right )$

where a = first term

r = common ratio

n = number of terms

## Examples on find common ratio when sum of n terms of geometric progression is given

1) Determine the common ratio and the number of terms in G.P. If $a_{1}$ = 3, $a_{n}$ = 96 and $S_{n} $= 189**Solution :**Let 'r' be the common ratio of the given G.P. then

$a_{n} = 96 ⇒ a_{1} r^{n - 1}$ = 96

⇒ 3$r^{n - 1}$ = 96

$r^{n - 1}$ = 32 ---------(1)

Now, $S_{n} $= 189

$S_{n} = a_{1}\left ( \frac{r^{n} - 1}{r - 1} \right )$

$a_{1}\left ( \frac{r^{n} - 1}{r - 1} \right )$ = 189

$r^{n} = r^{n -1} \times$ r

So, 3$\left ( \frac{(r^{n-1})r - 1}{r - 1} \right )$ = 189

but $r^{n -1}$= 32 and $a_{1}$ = 3

∴ 3$\left ( \frac{32r - 1}{r - 1} \right )$ = 189

$\left ( \frac{32r - 1}{r - 1} \right )$ = 63

32r -1 = 63(r -1)

32r - 1 = 63r -63

32r - 63r = -63 + 1

-31r = -62

∴ r = 2

Put r = 2 in equation (1) we get,

$2^{n - 1}$ = 32

$2^{n - 1} = 2^{5}$

∴ n - 1 = 5

⇒ n = 6

So, common ratio is 2 and number of terms are 6.

2) Find the common ratio of G.P., if first term is 2, last term is 486 and the sum of these terms be 728.

**Solution :**Let 'r' be the common ratio of the given G.P. then

$a_{n} = 486 ⇒ a_{1} r^{n - 1}$ = 486

⇒ 2$r^{n - 1}$ = 486

$r^{n - 1}$ = 243 ---------(1)

Now, $S_{n} $= 728

$S_{n} = a_{1}\left ( \frac{r^{n} - 1}{r - 1} \right )$

$a_{1}\left ( \frac{r^{n} - 1}{r - 1} \right )$ = 728

$r^{n} = r^{n -1} \times$ r

So, $a_{1}\left ( \frac{(r^{n-1})r - 1}{r - 1} \right )$ = 728

but $r^{n -1}$= 243 and $a_{1}$ = 2

2$\left ( \frac{(243r - 1}{r - 1} \right )$ = 728

$\left ( \frac{(243r - 1}{r - 1} \right )$ = 364

364r -364 = 243r -1

364r - 243r = -1 + 364

121r = 363

∴ r = 3

So common ratio is 3

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