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) $1, \frac{21}{11}, \frac{31}{11}, \frac{41}{11}, \ldots, x=\frac{171}{11}$
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 $n=17$.