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CBSE Class 10 Answered

if in an A.P. the sum of m terms equal to n and the sum of n terms is equal to m, then prove that the sum of (m+n) terms is -(m+n)
Asked by ananya_monica | 28 Sep, 2023, 07:07: PM
answered-by-expert Expert Answer
Let  a be the first term and d be the common difference of A.P.
 
Sum of m terms of  A.P
 
begin mathsize 14px style S subscript m space equals space m over 2 left square bracket space 2 a space plus space left parenthesis m minus 1 right parenthesis d space right square bracket space equals space n end style
begin mathsize 14px style left square bracket space 2 space a space plus space left parenthesis space m space minus 1 space right parenthesis space d space right square bracket space equals space fraction numerator 2 n over denominator m end fraction end style  ..................................(1)
 
Similarly , sum of n terms of A.P
 
begin mathsize 14px style left square bracket space 2 space a space plus space left parenthesis space n space minus 1 space right parenthesis space d space right square bracket space equals space fraction numerator 2 m over denominator n end fraction end style ............................... (2)
By solving eqn.(1) and (2) , we get
 
begin mathsize 14px style d space equals space open parentheses fraction numerator negative 2 over denominator m space n end fraction close parentheses space left parenthesis space m space plus space n space right parenthesis end style  ................................... (3)
begin mathsize 14px style a space equals space n over m plus space left parenthesis space m space minus 1 space right parenthesis space fraction numerator left parenthesis m plus n right parenthesis over denominator m n end fraction end style .........................(4)
 
Sum of (m+n) terms of A.P is
 
begin mathsize 14px style S subscript left parenthesis m plus n right parenthesis end subscript space equals space fraction numerator left parenthesis m plus n right parenthesis over denominator 2 end fraction left square bracket space 2 space a space plus space left parenthesis space m space plus space n space minus 1 space right parenthesis space d space right square bracket space end style ....................... (5)
By substituting initial value a from eqn.(4) and d from eqn.(3) , we get from eqn.(5) after simplification is
 
begin mathsize 14px style S subscript left parenthesis m plus n right parenthesis end subscript space equals space fraction numerator left parenthesis m plus n right parenthesis over denominator m space n end fraction open square brackets n squared plus left parenthesis m minus 1 right parenthesis left parenthesis m plus n right parenthesis space space minus space left parenthesis m plus n minus 1 right parenthesis space left parenthesis m plus n right parenthesis close square brackets end style
Above expression is simplified to
 
begin mathsize 14px style S subscript left parenthesis m plus n right parenthesis end subscript space equals space minus space left parenthesis space m space plus space n space right parenthesis end style
-------------------------------------


 
Answered by Thiyagarajan K | 28 Sep, 2023, 09:44: PM

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