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, 19:07: PM
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](https://images.topperlearning.com/topper/tinymce/cache/09c124e2160011802f32dcf4b49637e7.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/8744d7752f953f7866e0ffb511d49305.png)
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](https://images.topperlearning.com/topper/tinymce/cache/b4d9cbdb71dd68dafef7ee370151df0a.png)
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](https://images.topperlearning.com/topper/tinymce/cache/a8bcb61985b7fd93be6497ff7f2f60a3.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/9f88af6f051b0ed2289a3bd06da357b3.png)
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](https://images.topperlearning.com/topper/tinymce/cache/0627477f430135b0a8ff44c78e155a42.png)
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](https://images.topperlearning.com/topper/tinymce/cache/92cac990c2c9a4439cf4fa083a06292f.png)
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](https://images.topperlearning.com/topper/tinymce/cache/9efb187f3446affa628e6e3d01c8a7af.png)
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Answered by Thiyagarajan K | 28 Sep, 2023, 21:44: PM
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