# Find a30 − a20 for the A.P.

Question:

Find $a_{30}-a_{20}$ for the A.P.

(i) $-9,-14,-19,-24, \ldots$

(ii) $a, a+d, a+2 d, a+3 d, \ldots$

Solution:

In this problem, we are given different A.P. and we need to find $a_{30}-a_{20}$.

(i) A.P. $-9,-14,-19,-24, \ldots$

Here,

First term (a) = -9

Common difference of the A.P. $(d)=-14-(-9)$

$=-14+9$

$=-5$

Now, as we know,

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

Here, we find $a_{30}$ and $a_{20}$

So, for $30^{\text {th }}$ term,

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

$=-9+(29)(-5)$

$=-9-145$

$=-154$

Also, for $20^{\text {th }}$ term,

$a_{3}=a+(20-1) d$

$=-9+(19)(-5)$

$=-9-95$

$=-104$

So,

$a_{30}-a_{0}=-154-(-104)$

$=-154+104$

$=-50$

Therefore, for the given A.P $a_{30}-a_{20}=-50$

(ii) A.P. $a, a+d, a+2 d, a+3 d, \ldots$

Here,

First term $(a)=a$

Common difference of the A.P. $(d)=a+d-a=d$

Now, as we know,

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

Here, we find $a_{30}$ and $a_{20}$.

So, for $30^{\text {th }}$ term,

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

$=a+(29) d$

Also, for 20th term,

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

$=a+(19) d$

So,

$a_{30}-a_{20}=(a+29 d)-(a+19 d)$

$=a+29 d-a-19 d$

$=10 d$

Therefore, for the given A.P $a_{30}-a_{20}=10 d$

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