Question

(i) How many terms are there in the A.P.
7, 10, 13, ... 43?
(ii) How many terms are there in the A.P.
$-1,-\frac{5}{6},-\frac{2}{3},-\frac{1}{2},...,\frac{10}{3}?$

Answer 1

(i) 7, 10, 13...43
Here, we have:
a = 7
$$\begin{gathered} d=(10-7)=3 \\ {a}_n=43 \end{gathered}$$
Let there be n terms in the given A.P.

$$\begin{gathered} \mathrm{Also},a_n=a+\left(n-1\right)d \\ \Rightarrow 43=7+\left(n-1\right)3 \\ \Rightarrow 36=\left(n-1\right)3 \\ \Rightarrow 12=\left(n-1\right) \\ \Rightarrow 13=n \end{gathered}$$

Thus, there are 13 terms in the given A.P.

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

Here, we have:
a = $-1$
$$\begin{gathered} d=\left(\frac{-5}{6}-\left(-1\right)\right)=\left(1-\frac{5}{6}\right)=\frac{1}{6} \\ {a}_n=\frac{10}{3} \end{gathered}$$
Let there be n terms in the given A.P.

$$\begin{gathered} \mathrm{Also},a_n=a+\left(n-1\right)d \\ \Rightarrow \frac{10}{3}=-1+\left(n-1\right)\frac{1}{6} \\ \Rightarrow \frac{13}{3}=\left(n-1\right)\frac{1}{6} \\ \Rightarrow 26=\left(n-1\right) \\ \Rightarrow 27=n \end{gathered}$$

Thus, there are 27 terms in the given A.P.