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Permutations And Combinations

Question
CBSEENMA11013070

If the first and the nth terms of a GP. are a and b respectively and if P is the product of first n terms, prove that P2 = (ab)n.

Solution

Here, a is the first term, let r be the common ratio.
Also,       straight t subscript straight n space equals space straight b space space space rightwards double arrow space space ar to the power of straight n minus 1 end exponent space equals space straight b
rightwards double arrow      WiredFaculty
Since P is the product of first n terms

∴             straight P space equals space straight t subscript 1. space straight t subscript 2. space straight t subscript 3 space................ space straight t subscript straight n space equals space straight a. ar. ar squared. space............... space space space ar to the power of straight n minus 1 end exponent   open curly brackets because space space In space straight A. straight P. space straight S subscript straight n space equals space straight n over 2 left square bracket straight a plus straight l right square bracket close curly brackets
                  equals space straight a to the power of straight n straight r to the power of 1 plus 2 plus.......... plus space left parenthesis straight n space minus space 1 right parenthesis end exponent space equals space straight a to the power of straight n space. straight r to the power of fraction numerator straight n minus 1 over denominator 2 end fraction left parenthesis 1 plus straight n minus 1 right parenthesis end exponent
                  WiredFaculty
                  equals straight a to the power of straight n open square brackets open parentheses straight b over straight a close parentheses to the power of fraction numerator 1 over denominator straight n minus 1 end fraction end exponent close square brackets to the power of fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction end exponent space equals space straight a to the power of straight n open parentheses straight b over straight a close parentheses to the power of straight n divided by 2 end exponent space equals space straight a to the power of straight n. space straight b to the power of straight n divided by 2 end exponent over straight a to the power of straight n divided by 2 end exponent space equals space straight a to the power of straight n minus straight n over 2 end exponent straight b to the power of straight n over 2 end exponent equals space straight a to the power of straight n over 2 end exponent straight b to the power of straight n over 2 end exponent
rightwards double arrow              space space straight P space equals space left parenthesis ab right parenthesis to the power of straight n over 2 end exponent
Squaring both sides, we get WiredFaculty

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