If the 1st 2nd and last terms of an A.P are a, b and c, respectively

1.	If the 1st 2nd and last terms of an A.P are a, b and c, respectively then find the sum of all. terms of the A. P
Answer» Given A.P. is a, b … c First term (A) = a Common difference (d) = b − a Last term i.e., n^th term (A_n) = c Using formula, A_n = A + (n − 1)d, we get c = a + (n − 1)(b − a) ⇒ c − a + (n − 1)(b − a) ⇒ n − 1 = \(\frac{c−a}{b−a}\) ⇒ n = \(\frac{c−a}{b−a}\) + 1 ⇒ n = \(\frac{b+c−2a}{b−a}\) …...(i) Now, using formula S_n = \(\frac{n}{2}\)(A + A_n) for sum of n terms, we get S_n = \(\frac{n}{2}\)[a + c] = \(\frac{(b+c−2a)(a+c)}{2}\) using (i) = \(\frac{(a+c)(b+c−2a)}{2}\)

If the 1st 2nd and last terms of an A.P are a, b and c, respectively then find the sum of all. terms of the A. P

Answer»

Given A.P. is

a, b … c

First term (A) = a

Common difference (d) = b − a

Last term i.e., n^th term (A_n) = c

Using formula, A_n = A + (n − 1)d, we get

c = a + (n − 1)(b − a)

⇒ c − a + (n − 1)(b − a)

⇒ n − 1 = \(\frac{c−a}{b−a}\)

⇒ n = \(\frac{c−a}{b−a}\) + 1

⇒ n = \(\frac{b+c−2a}{b−a}\) …...(i)

Now, using formula S_n = \(\frac{n}{2}\)(A + A_n) for sum of n terms, we get

S_n = \(\frac{n}{2}\)[a + c]

= \(\frac{(b+c−2a)(a+c)}{2}\) using (i)

= \(\frac{(a+c)(b+c−2a)}{2}\)