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Question
For a G.P., if t4 = 16, t9 = 512, find S10.
Solution
t4 = 16, t9 = 512
tn = arn–1
∴ t4 = ar4–1 = 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, Sn = `("a"("r"^"n" - 1))/("r" - 1)` for r > 1
∴ S10 = `(2(2^10 - 1))/(2- 1)`
= 2(1024 – 1)
= 2046
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