Question

For a G.P., if t₃ = 20, t₆ = 160, find S₇.

Answer 1

t₃ = 20, t₆ = 160
t_n = ar^n–1
∴ t₃ = ar^3–1 = ar²
∴ ar² = 20

∴ a = `20/"r"^2`             ...(i)

Also, t₆ = ar⁵
∴ ar⁵ = 160

∴ `(20/"r"^2)"r"^5` = 160     ...[From (i)]

∴ r³ = `160/20` = 8

∴ r = 2
Substituting the value of r in (i), we get

a = `20/2^2` = 5

Now, S_n = `("a"("r"^"n" - 1))/("r" - 1)`, for > 1

∴ S₇ = `(5(2^7 - 1))/(2 - 1)`

= 5(128 – 1)
= 635.