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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))`.
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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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