Question

For a given AP the first term is 5 and the 19th term is 95. Find the sum of its 19 terms.​

Answer 1

GIVEN :-

• $\sf 1^{st} \ term = 5$

• $\sf 19^{th} \ term = 95$

TO FIND :-

• Sum of 19 terms of the AP.

SOLUTION :-

We know that,

$\bf a_n = a+ (n-1)d$

First term = 5

$\longrightarrow \sf a = 5$

$\bullet \sf \ a_{19}= 95$

$\longrightarrow \sf a+(19-1)d = 95 \\\\\longrightarrow a+18 d = 95 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .............(1)$

Substitute the value of a in equation(1),

$\longrightarrow \sf 5+18d= 95 \\\\\longrightarrow 18d = 95-5 \\\\\longrightarrow 18d = 90 \\\\\longrightarrow \bf d = 1$

$\boxed{\bf Sum \ of \ first \ n \ terms = \dfrac{n}{2}\times [ \ 2a+(n-1)d \ ]}$

 $\bullet \sf \ Sum \ of \ first \ 19 \ terms = \dfrac{19}{2} \times [ \ (2\times 5)+(19-1)\times 5 \ ]$

                                     $= \sf \dfrac{19}{2} \times [ \ 10+ 18\times 5 \ ]$

                                    $= \sf \dfrac{19}{2} \times( 10+90) \\\\= \dfrac{19}{2}\times 100 \\\\= \bf 950$