If first term of an A.P. is $a$ , second term is $b$ and last term is $c$ , then show that sum of all terms is $\frac{(a + c) (b + c - 2 a)}{2 (b - a)}$ .

Open in App

Solution

$\Rightarrow$ $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$\Rightarrow$ $b = a + d,$ $b - a = d$

$\Rightarrow$ $c = a + (n - 1) d$

$\Rightarrow$ $c = a + n d - d$

$\Rightarrow$ $c - a = (n - 1) (b - a)$

$\Rightarrow$ $(n - 1) = \frac{c - a}{b - a}, n = \frac{c - a}{b - a} + 1 = \frac{c - a + b - a}{b - a} = \frac{c + b - 2 a}{b - a}$

$\Rightarrow$ $d = (b - a)$

$\Rightarrow$ $S_{n} = \frac{c + b - 2 a}{2 (b - a)} [2 a + \frac{c - a}{b - a} (b - a)]$

$\Rightarrow$ $S_{n} = \frac{b + c - 2 a}{2 (b - a)} (a + c)$

$\Rightarrow$ $S_{n} = \frac{(a + c) (b + c - 2 a)}{2 (b - a)}$ .

Suggest Corrections

Similar questions

a

b

c

\frac{(a + c) (b + c - 2 a)}{2 (b - a)}

Show that the sum of a A.P. whose first term is 'a', the second term is 'b' and the last term is 'c', is equal to (a+c)(b+c-2a)/2(b-a)

\frac{(a + c) (b + c + 2 a)}{2 (b - a)}

a

b

c

\frac{(a + c) (b + c - 2 a)}{2 (b - a)}

\frac{(a + c) (b + c - 2 a)}{2 (b - a)}

Join BYJU'S Learning Program