Question

If the 3rd and 6th term of Arithmetic progression are 7and 13 respectively work out the sum of the first 20 term of the series

Answer 1

Solution :-

Let,

First term = a

Common difference = d

We know,

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

$\bullet \sf \ 3^{rd} \ term = 7$

 $\longrightarrow \sf a+(3-1)d = 7 \\\\\longrightarrow a+2d = 7 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......................(1)$

$\bullet \sf \ 6^{th} \ term = 13$

 $\longrightarrow\sf a+(6-1)d = 13 \\\\\longrightarrow a+5d = 13 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......................(2)$

Eq (2) - Eq (1),

$\longrightarrow\sf 3d = 6 \\\\\longrightarrow\bf d = 2$

Substitute the value of d in eq (1),

$\longrightarrow\sf a+2\times 2 = 7 \\\\\longrightarrow a+ 4 = 7 \\\\\longrightarrow \bf a = 3$

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

$\longrightarrow \sf Sum \ of \ first \ 20 \ terms = \dfrac{20}{2} \times [ \ 2\times 3 +(20-1) \times 2 \ ]$

                                       $= \sf 10 \times [ \ 6+19\times 2 \ ] \\\\= 10 \times [ \ 6 + 38 \ ] \\\\= 10 \times 44\\\\= 440$