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CBSE Class 10 Answered

If the sum of m terms of an AP is equal to the sum of either the next n terms or the next p terms of the same AP prove that (m+ n) [(1/m)-(1/p)] = (m + p) [(1/m) -(1/n)] .
Asked by rushabhjain.avv | 20 Feb, 2019, 09:02: PM
answered-by-expert Expert Answer
begin mathsize 16px style straight S subscript straight m equals straight m over 2 open square brackets 2 straight a plus left parenthesis straight m minus 1 right parenthesis straight d close square brackets straight S subscript straight n apostrophe equals fraction numerator straight m plus straight n over denominator 2 end fraction open square brackets 2 straight a plus open parentheses straight m plus straight n minus 1 close parentheses straight d close square brackets minus straight m over 2 open square brackets 2 straight a plus left parenthesis straight m minus 1 right parenthesis straight d close square brackets straight S subscript straight p apostrophe equals fraction numerator straight m plus straight p over denominator 2 end fraction open square brackets 2 straight a plus open parentheses straight m plus straight p minus 1 close parentheses straight d close square brackets minus straight m over 2 open square brackets 2 straight a plus left parenthesis straight m minus 1 right parenthesis straight d close square brackets According space to space the space question comma straight m over 2 open square brackets 2 straight a plus left parenthesis straight m minus 1 right parenthesis straight d close square brackets equals fraction numerator straight m plus straight n over denominator 2 end fraction open square brackets 2 straight a plus open parentheses straight m plus straight n minus 1 close parentheses straight d close square brackets minus straight m over 2 open square brackets 2 straight a plus left parenthesis straight m minus 1 right parenthesis straight d close square brackets Simplifying space we space get fraction numerator straight m minus straight n over denominator straight m plus straight n end fraction equals fraction numerator nd over denominator 2 straight a plus left parenthesis straight m minus 1 right parenthesis straight d end fraction.... left parenthesis straight i right parenthesis fraction numerator straight m minus straight p over denominator straight m plus straight p end fraction equals fraction numerator pd over denominator 2 straight a plus left parenthesis straight m minus 1 right parenthesis straight d end fraction... left parenthesis ii right parenthesis Dividing space left parenthesis ii right parenthesis space by space left parenthesis straight i right parenthesis and space simplifying space we space get space the space required space result. end style
Answered by Sneha shidid | 22 Feb, 2019, 11:08: AM

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