**Question:**

Find:

(i) Which term of the A.P. $3,8,13, \ldots$ is $248 ?$

(ii) Which term of the A.P. $84,80,76, \ldots$ is $248 ?$

(iii) Which term of the A.P. $4,9,14, \ldots$ is 254 ?

(iv) Which term of the A.P. $21,42,63,84, \ldots$ is 420 ?

(v) Which term of the A.P. $121,117,113, \ldots$ is its first negative term?

**Solution:**

In the given problem, we are given an A.P and the value of one of its term.

We need to find which term it is (*n*)

So here we will find the value of $n$ using the formula, $a_{n}=a+(n-1) d$

(i) Here, A.P is $3,8,13, \ldots$

$a_{n}=248$

$a=3$

Now,

Common difference $(d)=a_{1}-a$

$=8-3$

$=5$

Thus, using the above mentioned formula

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

$248=3+(n-1) 5$

$248-3=5 n-5$

$245+5=5 n$

$n=\frac{250}{5}$

$n=50$

Thus, $n=50$

Therefore 248 is the $50^{\text {th }}$ term of the given A.P

(ii) Here, A.P is $84,80,76, \ldots$

$a_{n}=0$

$a=84$

Now,

Common difference $(d)=a_{1}-a$

$=80-84$

$=-4$

Thus, using the above mentioned formula

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

$0=84+(n-1)(-4)$

$0=84-4 n+4$

$0=88-4 n$

$4 n=88$

On further simplifying, we get,

$n=\frac{88}{4}$

$n=22$

Thus, $n=22$

Therefore 84 is the $22^{\text {nd }}$ term of the given A.P

(iii) Here, A.P is $4,9,14, \ldots .$

$a_{n}=254$

$a=4$

Now,

Common difference $(d)=a_{1}-a$

$=9-4$

$=5$

Thus, using the above mentioned formula

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

$254=4+(n-1) 5$

$254-4=5 n-5$

$250+5=5 n$

$n=\frac{255}{5}$

$n=51$

Thus, $n=51$

Therefore 254 is the $51^{\text {t }}$ term of the given A.P

(iv) Here, A.P is $21,42,63,84, \ldots .$

$a_{n}=420$

$a=21$

Now,

Common difference $(d)=a_{1}-a$

$=42-21$

$=21$

Thus, using the above mentioned formula

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

$420=21+(n-1) 21$

$420-21=21 n-21$

$399+21=21 n$

$n=\frac{420}{21}$

$n=20$

Thus, $n=20$

Therefore 420 is the $20^{\text {th }}$ term of the given A.P

(v) Here, A.P is $121,117,113, \ldots .$

We need to find first negative term of the A.P

$a=121$

Now,

Common difference $(d)=a_{1}-a$

$=117-121$

$=-4$

Now, we need to find the first negative term,

$a_{n}<0$

$121+(n-1)(-4)<0$

$121-4 n+4<0$

$125-4 n<0$

$4 n>125$

Further simplifying, we get,

$n>\frac{125}{4}$

$n>31 \frac{1}{4}$

$n \geq 32$ (as $n$ is a natural number)

Thus, $n=32$

Therefore, the first negative term is the $32^{\text {nt }}$ term of the given A.P.