Question

If the first of an A.P. is a, second term is b and last term is c, then show that sum of all terms is [(a+c)(b+c-2a)]/2(b-a).

Answer 1

The answer is given below :

The first term of the AP = a

and the common difference, d = (b - a)

Let, the AP has n number of terms in total.

Then, the last term (nth term)

c = a + (n - 1)d

⇒ c = a + (n - 1) (b - a)

⇒ 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)

⇒ n = (b + c - 2a)/(b - a)

So, the given AP has (b + c - 2a)/(b - a) number of terms.

Now, the sum of all the n terms

= n/2 × (first term + last term)

= {(b + c - 2a)/(b - a)}/2 × (a + c), putting the value of n

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

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