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



11th grade math

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