Question

The third term of an A.P is 4m - 2n. If the ninth term of the progression is 2m - 8n. Find the common difference in terms of m and n​

Answer 1

Let $a_n$ denote the n-th term in the progression. So

$a_n=a_{n-1}+d$

for some constant difference between terms d.

Solve for $a_n$ explicitly:

$a_4=a_3+d$

$a_5=a_4+d=a_3+2d$

$a_6=a_5+d=a_3+3d$

and so on, up to

$a_n=a_3+(n-3)d$

We're told that the third term is $a_3=4m-2n$, and the ninth term is $a_9=2m-8n$, and according to the recursive rule above, we have

$a_9=a_3+6d$

Solve for d :

$2m-8n=(4m-2n)+6d$

$-2m-6n=6d$

$d=-\dfrac{2m+6n}6=\boxed{-\dfrac{m+3n}3}$