Question

the sum of 4th and 6th term of an ap is 75 and the sum of 5th and 10th terms of an ap is 265 find the AP​

Answer 1

$\bf{\Huge{\boxed{\bf{\green{ANSWER\::}}}}}$

$\bf{\Large{\underline{\tt{Given\::}}}}}$

The sum of 4th & 6th term of an A.P. is 75 & the sum of 5th & 10th term of an A.P. is 265.

$\bf{\Large{\underline{\tt{To\:\:find\::}}}}$

The A.P.

$\bf{\Large{\underline{\sf{\red{Explanation\::}}}}}$

We know that formula of the terms:

$\leadsto\tt{\orange{an=a+(n-1)d}}$

A/Q

$\longmapsto\tt{a4\:+\:a6\:=\:75}$

$\longmapsto\tt{a+(4-1)d+a+(6-1)d=75}$

$\longmapsto\tt{a+3d+a+5d\:=\:75}$

$\longmapsto\tt{2a+8d\:=\:75.....................(1)}$

&

$\longmapsto\tt{a5\:+\:a10\:=\:265}$

$\longmapsto\tt{a+(5-1)d+a+(10-1)d\:=\:265}$

$\longmapsto\tt{a+4d+a+9d=265}$

$\longmapsto\tt{2a+13d\:=\:265.........................(2)}$

Subtracting equation (1) from equation (2), we get;

$\longmapsto\tt{2a+13d-(2a+8d)=265-75}$

$\longmapsto\tt{\cancel{2a}+13d\cancel{-2a}-8d=265-75}$

$\longmapsto\tt{5d\:=\:190}$

$\longmapsto\tt{d\:=\:\cancel{\frac{190}{5} }}$

$\longmapsto\tt{\orange{d\:=\:38}}$

Putting the value of d in equation (1), we get;

$\longmapsto\tt{2a+8(38)=75}$

$\longmapsto\tt{2a+304\:=\:75}$

$\longmapsto\tt{2a\:=\:75\:-\:304}$

$\longmapsto\tt{2a\:=\:-229}$

$\longmapsto\tt{\orange{a\:=\:-\frac{229}{2} }}$

Thus,

The Arithmetic Progression is -229/2,-153/2 , -77/2.............