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# Circled one ques

Asked by rbhatt16 20th October 2017, 10:02 AM
Though the solution to the question is worked out, this question is posted indicates
the student needs to understand intermediate steps.

The above series in Arithmetic progression with the first term ' a = 1' and common difference ' d=2';

Hence in first series last term p means (2n-1) = p hence n = (p+1)/2

first series has (p+1)/2 terms and their sum is
Similarly second series has (q+1)/2 terms and their sum is
and Third series has (r+1)/2 terms and their sum is

since p>6, a >7

smallest phthagorean triplet containing one side number greater than 7 is  (6,8,10) and is given by
Hence we can consider a = 8, b = 6 and c = 10. This gives p = 7, q = 5; r = 9

hence p+q+r = 7+5+9 = 21

Answered by Expert 15th December 2017, 11:08 AM
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