Question

The sum of first 3 terms of a G.P is 16

and the sum of next three terms is 128.

Find the first term, common ratio and

sum of n terms.​

Answer 1

Step-by-step explanation:

Let a be the first term and r the common ratio of the G.P.

a+ar+ar

2

=16

and, ar

3

+ar

4

+ar

5

=128

a(1+r+r

2

)=16 and, ar

3

(1+r+r

2

)=128

⇒

a(1+r+r

2

)

ar

3

(1+r+r

2

)

=

16

128

⇒r

3

=8

⇒r=2

Putting the value of r, we get, a=

7

16

Therefore,

S

n

=a(

r−1

r

n

−1

)=

7

16

(

2−1

2

n

−1

)=

7

16

(2

n

−1)

Answer 2

Let a be the first term and r the common ratio of the G.P. 

a + ar + ar² = 16

and, ar³ + ar⁴ +ar^5=128

a(1+r+r²)=16 and, ar³(1+r+r²)=128

$\large\red{=> \frac{ {ar}^{3(1 + r + \frac{r}{2} )} }{a(1 + r + {r}^{2} )} = \frac{128}{16}}$

⇒r³=8

⇒r=2

Putting the value of r, we get, $a=\frac{16}{7}$

Therefore, 

$\large\pink{sn = a( \frac{ {r}^{n} - 1}{r - 1} ) = \frac{16}{7} ( \frac{ {2}^{n} - 1}{2 - 1} ) = \frac{16}{7} ( {2}^{n} - 1)}$