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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 G.P. be aarar2ar3, …

According to the given condition,

ar + ar2 = 16 and ar3 + ar4 + ar5 = 128

⇒ a (1 + r + r2) = 16 … (1)

ar3(1 + r + r2) = 128 … (2)

Dividing equation (2) by (1), we obtain

fraction numerator a r cubed open parentheses 1 plus r plus r squared close parentheses over denominator a open parentheses 1 plus r plus r squared close parentheses end fraction space equals space 128 over 16rightwards double arrow space r cubed space equals space 8therefore space r space equals space 2

Substituting r = 2 in (1), we obtain

a (1 + 2 + 4) = 16

⇒ a (7) = 16

rightwards double arrow a space equals space 16 over 7S subscript n space end subscript equals space fraction numerator a open parentheses r to the power of n space minus 1 close parentheses over denominator r minus 1 end fractionrightwards double arrow S subscript n space equals space fraction numerator 16 space over denominator 7 end fraction fraction numerator open parentheses 2 to the power of n space minus 1 close parentheses over denominator 2 minus 1 end fraction space equals space 16 over 7 space open parentheses 2 to the power of n space minus 1 close parentheses

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