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Worked Examples · Example 12

Q.If the first, second and last terms of an A.P. are aa, bb and cc respectively, then show that the sum of the terms in the A.P. is (a+c)(b+c−2a)2(b−a)\dfrac{(a+c)(b+c-2a)}{2(b-a)}.

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Finding nn from the last-term relation c=a+(n−1)dc=a+(n-1)d and substituting into Sn=n2(a+c)S_n=\frac{n}{2}(a+c) produces exactly the required expression.

For an A.P. with first term aa, common difference dd, and nn terms whose last term is ll:

l=a+(n−1)d,Sn=n2(a+l)l = a+(n-1)d, \qquad S_n = \frac{n}{2}(a+l)

Here the given first, second, and last terms are aa, bb, cc respectively, so d=b−ad=b-a and l=cl=c.

  1. The common difference is d=b−ad = b-a (difference between the 2nd and 1st terms).
  2. The last term is given as cc, so c=a+(n−1)d=a+(n−1)(b−a)c = a+(n-1)d = a+(n-1)(b-a).
  3. Solve for n−1n-1: n−1=c−ab−an-1 = \dfrac{c-a}{b-a}.
  4. So n=1+c−ab−a=(b−a)+(c−a)b−a=b+c−2ab−an = 1+\dfrac{c-a}{b-a} = \dfrac{(b-a)+(c-a)}{b-a} = \dfrac{b+c-2a}{b-a}.
  5. Now use the sum formula with last term l=cl=c: Sn=n2(a+c)S_n = \dfrac{n}{2}(a+c). …

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