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Question
How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?
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Solution
Let the total terms of the geometric progression = n.
First term, a = 3, common ratio, r = `3^2/3 = 3`
Sum of n terms = `("a"("r"^"n" - 1))/("r" - 1), "r" >1`
= `(3(3^"n" - 1))/(3 - 1)`
= 120
or 3(3n – 1)
= 120 × 2
= 240
dividing by 3
3n – 1
= `240/3`
= 80
Or 3n = 80 + 1 = 81 = 34
∴ n = 4
Hence, 4 terms are needed.
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