Question

For a G.P., if t₄ = 16, t₉ = 512, find S₁₀.

Answer 1

t₄ = 16, t₉ = 512
t_n = ar^n–1
∴ t₄ = ar^4–1 = ar³

∴ a = `16/"r"^3`           ...(i)

Also, t₉ = ar⁸
∴ ar⁸ = 512

∴ `16/"r"^3 xx "r"^8` = 512

∴ r⁵ = 32
∴  r = 2
Substituting r = 2 in (i), we get

a = `16/2^3 = 16/8` = 2

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

∴ S₁₀ = `(2(2^10 - 1))/(2- 1)`

= 2(1024 – 1)
= 2046