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Home / Doubts and Solutions/JEE/Mathematics

Circled one ques

Asked by rbhatt16 20th October 2017, 10:02 AM
Answered by Expert
Answer:
Though the solution to the question is worked out, this question is posted indicates
the student needs to understand intermediate steps. 
 
begin mathsize 12px style L e t space u s space c o n s i d e r space P subscript n equals space 1 plus 3 plus 5 plus............. p semicolon end style
The above series in Arithmetic progression with the first term ' a = 1' and common difference ' d=2';
 
begin mathsize 12px style H e n c e space P subscript n space equals space n over 2 space left parenthesis 2 a space plus space left parenthesis n minus 1 right parenthesis d right parenthesis space equals space n over 2 left parenthesis 2 plus left parenthesis n minus 1 right parenthesis 2 right parenthesis space equals space n squared S u m space o f space n space t e r m s space equals space n squared end style
begin mathsize 12px style I n space t h e space a b o v e space s e r i e s space t h e space n to the power of t h end exponent space t e r m space i s space g i v e n space b y space a plus left parenthesis n minus 1 right parenthesis d space equals space 1 plus left parenthesis n minus 1 right parenthesis 2 space equals space 2 n minus 1 end style
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 begin mathsize 12px style open parentheses p plus 1 close parentheses squared over 4 end style
Similarly second series has (q+1)/2 terms and their sum is open parentheses q plus 1 close parentheses squared over 4
and Third series has (r+1)/2 terms and their sum is open parentheses r plus 1 close parentheses squared over 4
begin mathsize 12px style left parenthesis 1 plus 3 plus 5....... p right parenthesis space plus space left parenthesis 1 plus 3 plus 5....... q right parenthesis space plus left parenthesis 1 plus 3 plus 5....... r right parenthesis space space equals space open square brackets fraction numerator open parentheses p plus 1 close parentheses over denominator 2 end fraction close square brackets squared plus space open square brackets fraction numerator open parentheses q plus 1 close parentheses over denominator 2 end fraction close square brackets squared equals space open square brackets fraction numerator open parentheses r plus 1 close parentheses over denominator 2 end fraction close square brackets squared space open parentheses p plus 1 close parentheses squared space plus space open parentheses q plus 1 close parentheses squared space equals space open parentheses r plus 1 close parentheses squared l e t space a space equals space p plus 1 comma space b space equals space q plus 1 comma space c space equals space r plus 1 semicolon t h e n space a squared plus b squared space equals space c squared space semicolon space T h i s space i m p l e s space a comma b space a n d space c space a r e space p h y t h o g o r o u s space t r i p l e t s end style
 
since p>6, a >7
 
smallest phthagorean triplet containing one side number greater than 7 is  (6,8,10) and is given by begin mathsize 12px style 6 squared plus 8 squared space equals space 10 squared end style
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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