Question

The sum of first 3 terms of GP is 16 and sum of next 3 terms is 128 . find 1 st terms and commom ratio

Answer 1

let gp be a,ar,ar^2,ar^3………
according to given,
a+ar+ar^2=16........(1)
ar^3+ar^4+ar^5=128........(2)
=>a(1+r+r^2)=16
a^3(1+r+r^3)=128

on dividing eq.2 by 1
ar^3(1+r+r^3)=128
_____________
a(1+r+r^2)=16

=>r^3=8
=>r=2
on substituting r=2 in eq 1
a(1+2+4)=16
a=16/7

Sn=a(r^n-1)
______=>16/7(2^n-1)
r-1. _____
2-1
=>16/7(2^n-1)
answer

Answer 2

=> Let G.P. is a , ar , ar^2 , ar^3 ,....

According to question ,

=> $a + ar + {ar}^{2} = 16$ (1)

=> ${ar}^{3} + {ar}^{4} + {ar}^{5} = 128$(2)

Now Taking Common,

=> $a(1 + r + {r}^{2} ) = 16$

=> ${ar}^{3} (1 + r + {r}^{2} ) = 128$

Divide eq. 2 by eq. 1

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

=> ${r}^{3} = 8$

=> $r = 2$

Put r = 2 in eq. 1.

=> $a + a \times 2 + a \times {2}^{2} = 16$

$= > a + 2a + 4a = 16$

$= > 7a = 16$

$= > a = \frac{16}{7}$

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