# How many terms are there in the A.P.?

Question:

How many terms are there in the A.P.?

(i) $7,10,13, \ldots 43$.

(ii) $-1, \frac{5}{6}, \frac{2}{3}, \frac{1}{2}, \ldots \frac{10}{3}$.

(iii) $7,13,19, \ldots, 205$.

(iv) $18,15 \frac{1}{2}, 13, \ldots,-47$.

Solution:

In the given problem, we are given an A.P.

We need to find the number of terms present in it

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

(i) Here, A.P is $7,10,13, \ldots .43$

The first term $(a)=7$

The last term $\left(a_{n}\right)=43$

Now,

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

$=10-7$

$=3$

Thus, using the above mentioned formula, we get,

$43=7+(n-1) 3$

$43-7=3 n-3$

$36+3=3 n$

$n=\frac{39}{3}$

$n=13$

Thus, $n=13$

Therefore, the number of terms present in the given A.P is 13

(ii) Here, A.P is $-1,-\frac{5}{6},-\frac{2}{3},-\frac{1}{2}, \ldots, \frac{10}{3}$

The first term $(a)=-1$

The last term $\left(a_{n}\right)=\frac{10}{3}$

Now,

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

$=-\frac{5}{6}-(-1)$

$=-\frac{5}{6}+1$

$=\frac{-5+6}{6}$

$=\frac{1}{6}$

Thus, using the above mentioned formula, we get,

$\frac{10}{3}=-1+(n-1) \frac{1}{6}$

$\frac{10}{3}+1=\frac{1}{6} n-\frac{1}{6}$

$\frac{13}{3}+\frac{1}{6}=\frac{1}{6} n$

Further solving for n, we get

$\frac{26+1}{6}=\frac{1}{6} n$

$n=\frac{27}{6}(6)$

$n=27$

Thus, $n=27$

Therefore, the number of terms present in the given A.P is 27

(iii) Here, A.P is $7,13,19, \ldots .205$

The first term $(a)=7$

The last term $\left(a_{n}\right)=205$

Now,

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

$=13-7$

$=6$

Thus, using the above mentioned formula, we get,

$205=7+(n-1) 6$

$205-7=6 n-6$

$198+6=6 n$

$n=\frac{204}{6}$

$n=34$

Thus, $n=34$

Therefore, the number of terms present in the given A.P is 34

(iv) Here, A.P is $18,15 \frac{1}{2}, 13, \ldots,-47$

The first term $(a)=18$

The last term $\left(a_{n}\right)=-47$

Now,

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

$=15 \frac{1}{2}-18$

$=\frac{31}{2}-18$

$=\frac{31-36}{2}$

$=-\frac{5}{2}$

Thus, using the above mentioned formula, we get,

$-47=18+(n-1)\left(-\frac{5}{2}\right)$

$-47-18=-\frac{5}{2} n+\frac{5}{2}$

$-65-\frac{5}{2}=-\frac{5}{2} n$

Further, solving for n, we get

$\frac{-130-5}{2}=-\frac{5}{2} n$

$-\frac{135}{2}(2)=-5 n$

$n=\frac{-135}{-5}$

$n=27$

Thus, $n=27$

Therefore, the number of terms present in the given A.P is 27 .