Question 10

Find the sum to n terms in the geometric progression x3, x5, x7 ... (if x ≠ ±1)

Answer

The given G.P. is  x3, x5, x7 ........

Here, a = x3 and r = x2

S subscript n space end subscript equals space fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction equals fraction numerator x cubed open square brackets 1 minus open parentheses x squared close parentheses to the power of n close square brackets over denominator 1 minus x squared end fraction equals fraction numerator x cubed open parentheses 1 minus x to the power of 2 n end exponent close parentheses over denominator 1 minus x squared end fraction

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