Find n if the given value of x is the nth term of the given A.P.
Question:

Find n if the given value of x is the nth term of the given A.P.

(i) $25,50,75,100, \ldots, x=1000$

(ii) $-1,-3,-5,-7, \ldots ; x=-151$

(iii) $5 \frac{1}{2}, 11,16 \frac{1}{2}, 22, \ldots ; x=550$

(iv) $1, \frac{21}{11}, \frac{31}{11}, \frac{41}{11}, \ldots, x=\frac{171}{11}$

Solution:

In the given problem, we need to find the number of terms in an A.P

(i) 25, 50, 75, 100 …

We are given,

$a_{n}=1000$

Let us take the total number of terms as n.

So,

First term (a) = 25

Last term (an) = 1000

Common difference $(d)=50-25$

$=25$

Now, as we know,

$a_{0}=a+(n-1) d$

So, for the last term,

$1000=25+(n-1) 25$

$1000=25+25 n-25$

$1000=25 n$

$n=\frac{1000}{25}$

$n=40$

Therefore, the total number of terms of the given A.P. is $n=40$.

(ii) $-1,-3,-5,-7 \ldots$

We are given,

$a_{n}=-151$

Let us take the total number of terms as n.

So,

First term (a) = −1

Last term (an) = −151

Common difference $(d)=-3-(-1)$

$=-3+1$

$=-2$

Now, as we know,

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

So, for the last term,

$-151=-1+(n-1)(-2)$

$-151=-1-2 n+2$

$-151=1-2 n$

$-2 n=-|5|-1$

$-2 n=-151-1$

On further simplifying, we get,

$-2 n=-152$

$n=\frac{-152}{-2}$

$n=76$

Therefore, the total number of terms of the given A.P. is $n=76$.

(iii) $5 \frac{1}{2}, 11,16 \frac{1}{2}, 22, \ldots$

We are given,

$a_{s}=550$

Let us take the total number of terms as n.

So,

First term $(a)=5 \frac{1}{2}$

Last term $\left(a_{n}\right)=550$

Common difference $(d)=11-5 \frac{1}{2}$

$=11-\frac{11}{2}$

$=\frac{22-11}{2}$

$=\frac{11}{2}$

Now, as we know,

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

So, for the last term,

$550=5 \frac{1}{2}+(n-1)\left(\frac{11}{2}\right)$

$550=\frac{11}{2}+\frac{11}{2} n-\frac{11}{2}$

$550=\frac{11}{2} n$

$n=\frac{550(2)}{11}$

On further simplifying, we get,

$n=\frac{1100}{11}$

$n=100$

Therefore, the total number of terms of the given A.P. is $n=100$

(iv) $1, \frac{21}{11}, \frac{31}{11}, \frac{41}{11}, \ldots$

We are given,

$a_{n}=\frac{171}{11}$

Let us take the total number of terms as n.

So,

First term $(a)=1$

Last term $\left(a_{n}\right)=\frac{171}{11}$

Common difference $(d)=\frac{21}{11}-1$

$=\frac{21-11}{11}$

$=\frac{10}{11}$

Now, as we know,

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

So, for the last term,

$\frac{171}{11}=1+(n-1)\left(\frac{10}{11}\right)$

$\frac{171}{11}=1+\frac{10}{11} n-\frac{10}{11}$

$\frac{171}{11}=\frac{11-10}{11}+\frac{10}{11} n$

$\frac{171}{11}=\frac{1}{11}+\frac{10}{11} n$

On further simplifying, we get,

$\frac{10}{11} n=\frac{171}{11}-\frac{1}{11}$

$\frac{10}{11} n=\frac{170}{11}$

$n=\frac{(170)(11)}{(11)(10)}$

$n=17$

Therefore, the total number of terms of the given A.P. is $n=17$.