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The first term of an A.P. is a, the second term is b and the last term is c. Show that the sum of the A.P. is (b+c-2a)(c+a)2(b-a).

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Question

The first term of an A.P. is a, the second term is b and the last term is c. Show that the sum of the A.P. is `((b + c - 2a)(c + a))/(2(b - a))`.

Sum
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Solution

Let d be the common difference and n be the number of terms of the A.P.

Since the first term is a and the second term is b

Therefore, d = b – a

Also, the last term is c

So c = a + (n – 1)(b – a)   .....(Since d = b – a)

⇒ n – 1 = `(c - a)/(b - a)`

⇒ n = `1 + (c - a)/(b - a)`

= `(b - a + c - a)/(b - a)`

= `(b + c - 2a)/(b - a)`

Therefore, Sn = `n/2 (a + 1)`

= `((b + c - 2a))/(2(b - a)) (a + c)`

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Chapter 9: Sequences and Series - Solved Examples [Page 150]

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NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 9 Sequences and Series
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