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Suppose the sum of the dirst m teams of a arithmetic progression is n and the sum of its first n terms is m, where `mnen.` Then the dum of the first (m + n) terms of the arithmetic progression isA. 1- mnB. mn- 5C. `-(m = n)`D. m + n |
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Answer» Correct Answer - C `S_(m)=m/2[2a+(m-1)d]=n" …(1)"` `S_(n)=n/2[2a+(n-1)d]=m" …(2)"` by (1) and (2) `(m-n)a+(m-n){m+n-1}d/2= -(m-n)` `rArr2a+(m+n-1)d= -2(mnen)` `rArrS_(m+n)=(m+n)/2[2a+(m+n-1)d]= -(m+n)` |
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