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Sia ? 6 years, 6 months ago
Let there be n terms in the given A.P.
We have, First term = a, second term = b
{tex}\therefore{/tex} Common difference d = b - a
It is given that the last term is c i.e. nth term an = c.
an = a + (n - 1)d
{tex}\therefore{/tex} c = a + (n - 1)d
{tex}\Rightarrow{/tex} c = a + (n - 1) (b - a)
{tex}\Rightarrow{/tex} (c - a) = (n - 1) (b - a)
{tex}\Rightarrow \quad n - 1 = \frac { c - a } { b - a } {/tex}
{tex}\Rightarrow \quad n = \frac { c - a } { b - a } {/tex}+ 1
{tex}\Rightarrow n = \frac { c -a + b -a } { b - a }{/tex}
{tex}\Rightarrow n = \frac { b + c - 2 a } { b - a }{/tex}...(1)
Let Sn be the sum of n terms of the A.P.
Then,
{tex}S _ { n } = \frac { n } { 2 } ( a + c ) = \frac { ( b + c - 2 a ) ( a + c ) } { 2 ( b - a ) }{/tex}[Using (1)]
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