Question

the difference between first and fourth term of an geometric progression is 52 if the sum of its first three term is 26 then find the sum of first six terms​

Answer 1

Answer:

Sum of first six terms is 728

Step-by-step explanation:

Formula used:

Sum of n terms of a G.P $=\frac{a(r^n-1)}{r-1}$

Given:

$ar^3-a=52$...........(1)

and

$a+ar+ar^2=26$.......(2)

Now, (1) can be written as

$a(r^3-1^3)=52$

$a(r-1)(r^2+r+1)=52$

$(r-1)(ar^2+ar+a)=52$

$(r-1)(26)=52$ (using (2))

$r-1=\frac{52}{26}$

$r-1=2$

$r=3$

put r=3 in (1) we get

$a(3^3-1)=52$

$a(26)=52$

$a=\frac{52}{26}$

$a=2$

Sum of first six terms

$=S_6$

$=\frac{a(r^6-1)}{r-1}$

$=\frac{2(3^6-1)}{3-1}$

$=\frac{2(729-1)}{2}$

$=728$

Answer 2

Answer:

the difference between first and fourth term of an geometric progression is 52 if the sum of its first three term is 26 then find the sum of first six terms

728