Write the value of a30 − a10 for the A.P. 4, 9, 14, 19, ....

Solution:

In this problem, we are given an A.P. and we need to find.

A.P. is 

Here,

First term (a) = 4

Common difference of the A.P. (d) 

Now, as we know,

$a_{n}=a+(n-1) d$

Here, we find a30 and a20.

So, for 30th term,

$a_{30}=a+(30-1) d$

$=4+(29)(5)$

$=4+145$

 

$=149$

Also, for 10th term,

$a_{20}=a+(10-1) d$

$=4+(9)(5)$

$=4+45$

 

$=49$

So,

$a_{30}-a_{10}=149-49$

$=100$

Therefore, for the given A.P $a_{30}-a_{10}=100$.