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Question
The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
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Solution
Let the geometric progression be a, ar, ar2, …..
First term = a, common ratio = r
Sum of three terms = `("a"(1 - "r"^3))/(1 - "r") = 16` ..........(i)
Fourth term = a × rn -1 = ar4-1 = ar3
Sum of next three terms = `("ar"^3(1 - "r"^3))/(1 - "r") = 128` ........(ii)
Dividing equation (ii) by (i), we get
`("ar"^3(1 - "r"^3))/(1 - "r") xx (1 - "r")/("a"(1 - "r"^3))`
= `128/16`
= 8
∴ r3 = 8 or r = 2
∴ On keeping the value of r in equation (i)
`("a"(1 - 8))/(1 - 2) = 16` or 7a = 16
∴ a = `16/7`
Here r > 1
∴ Sn = `(16/7(2^"n" - 1))/(2 -1)`
= `16/7 (2^"n" - 1)`
Hence a = `16/7`, r = 2, Sn = `16/7(2"n" - 1)`
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