In an A.P. the first term is 2, the last term is 29 and the sum of the terms is 155. Find the common difference of the A.P.


Given: An A.P. with the first term $a=2$ and last term $l=2$ and the sum of the term $S=155$

To do: To find out the common difference of the given A.P.

Solution:
Here in the given A.P.

First term $a=2$ 

Last term $l=29$

Common difference $d=?$

Number of terms $n=?$

As we know sum of n terms in A.P. $=\frac{n}{2}( a+l)$

By putting the values of a, l and sum we get

$155=\frac{n}{2}( 2+29)$

$\Rightarrow 31n=310$

$\Rightarrow n=\frac{310}{31} =10$

Here we obtained the number of terms $n=10$

$n^{th}$ term of an A.P.$ =a+( n-1) d$

$\Rightarrow 29=2+( 10-1) d\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( as\ we\ know\ here\ n^{th} term\ \ is\ 29\ and\ n=10\right)$

$\Rightarrow 29=2+9d$

$\Rightarrow 9d=27$

$\Rightarrow d=\frac{27}{9} =3$

Common difference of the given A.P. $d=3$.

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Updated on: 10-Oct-2022

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