Question

Sum of m terms of an A.P is equal to the Sum of n terms of that A.P , Prove that, Sum of its (m+n) terms of A.P is 0

Answer 1

Sum of the first m terms of an AP = sum of it's first n http://terms.So

(m/2){2a+(m-1)d} = (n/2){2a+(n-1)d}

ie. m{2a+(m-1)d} = n{2a+(n-1)d}

ie. 2a(m-n) +( m^2-n^2)d -md +nd = 0

ie 2a(m-n) + (m+n)(m-n)d -d(m-n) = 0

ie. (m-n){2a + (m+n-1)d} = 0

Here m-n is not equal to 0. So

2a + (m+n-1)d =0

So sum of the first m+n terms

={(m+n)/2}{2a+(m+n-1)d}={(m+n)/2 × 0 = 0

Answer 2

hey mate,here is my perfect answer .


Let, the first term and common difference of the respective AP be a and dAgain, let,Sum of the first m-terms of the AP Sm,Sum of the first n-terms of the AP Sn& Sum of the first mn-terms of the AP SmnAccording to the question,SmSnm22am1dn22an1d2amm2md2ann2nd2amnm2n2mnd0mn2amn1d02amn1d0Assuming, mnNow,Smnmn22amn1dmn20From Smn0.Hope, you'll understand..!!P.S.- Feel free to comment if you have any query with it.Thank You!


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