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If the sum of first three terms of an ap is a and sum of last three terms is b. If n be the no. Of terms then show that the sum of all terms is given by Sn=n(a+b)/6
Asked by rushabhjain.avv | 18 Nov, 2018, 04:49: PM
If three numbers are in A.P , then middle number is one-third of sum of these three numbers.

This we can verify, if we assume the three numbers are a-d, a and a+d so that middle number is one-third of sum 3a.

Let A be the first number in A.P of n numbers and the common difference is d.

We are given that sum of first three numbers in A.P is a.

Hence 2nd term of A.P is given by,   A+d = a/3 ...................(1)

Similarly if sum last three numbers in A.P is b,

(n-1)th term is given by,  A+(n-2)d = b/3  ..................(2)

By subtracting eqn.(1) from eqn.(2),  and solving for d, we get

...........................(3)

Now sum S of A.P is given by ,     S = (n/2) [ 2×A + (n-1) d ] = (n/2) [ 2 (A+d) + (n-1)d - 2d ]  = (n/2)[ 2(A+d) +(n-3)d ] .................(4)

substituting in eqn.(4)  for (A+d) from eqn.(1) and for (n-3)d from eqn.(3), we get,   S = (n/2) [ 2a/3  +(b-a)/3 ] = n(a+b)/6
Answered by Thiyagarajan K | 19 Nov, 2018, 12:09: AM

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