Question

the sum of the first three terms of a GP is 16 and the sum of the next three terms is 128.Then the first term,common ratio and to n terms of the GP respectively are​

Answer 1

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

a + ar + ar² = 16                       ----- (1)

and,

ar³ + ar⁴ + ar⁵ = 128                 -----(2)

Dividing (2) by (1)

So, $\frac{ar^{3}(1+r+r^{2}) }{a(1+r+r^{2})} =\frac{128}{26}$

Or, $r^{3} =8$

Or, $r=2$

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

Therefore,  

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

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Answer 2

Step-by-step explanation:

let the GP be a/r ,a and , ar

so according to question ,

a/r + a + ar = 16 and ,. (1)

ar² + ar³ + ar⁴ =128. (2)

In eqn. 2, r²(a + ar + ar²) = 128

and. 1/r( a + ar + ar²)= 16

Dividing both the eqn.

r²(a + ar + ar²) = 128

1/r( a + ar + ar²)= 16

This will give result as ,

r³= 8 or r=2

Now the value of a becomes a/2 + a + 2a = 16

so , 7a/ 2 = 16

first term (a/2) = 16/7

Using the formula of sum of GP series ,

sum = 16/7(2^n -1)