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Question
For a G.P., if t3 = 20, t6 = 160, find S7.
Solution
t3 = 20, t6 = 160
tn = arn–1
∴ t3 = ar3–1 = 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, Sn = `("a"("r"^"n" - 1))/("r" - 1)`, for > 1
∴ S7 = `(5(2^7 - 1))/(2 - 1)`
= 5(128 – 1)
= 635.
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