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Sum

Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to `((a + c)(b + c - 2a))/(2(b - a))`

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#### Solution

Given that, the AP is a, b, , c

Here, first term = a, common difference = b – a

And last term, (l) = a_{n} = c

∵ a_{n} = l = a + (n – 1 )d

⇒ c = a + (n – 1)(b – a)

⇒ `(n - 1) = (c - a)/(b - a)`

`n = (c - a)/(b - a) + 1`

⇒ `n = (c - a + b - a)/(b - a)`

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

∴ Sum of an AP,

`S_n = n/2 [2a + (n - 1)d]`

= `((b + c - 2a))/(2(b - a)) [2a + {(b + c - 2a)/(b - a) - 1}(b - a)]`

= `((b + c - 2a))/(2(b - a))[2a + (c - a)/(b - a) * (b - a)]`

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

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

Hence proved.

Concept: Sum of First n Terms of an A.P.

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