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. $-1,\frac{-5}{6}-\frac{2}{3},\frac{1}{2},\dots \dots \frac{10}{3}$ ?

Answer 1

(i) The given A.P. is 7, 10, 13, ....... 43.

a = 7

d = (10 - 7) =3 $a_n=43$

Let there be n terms in the given A.P.

$a_n=a+(n-1)d$

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

$\Rightarrow 36=(n-1{)}^3$

$\Rightarrow 12=(n-1)$

$\Rightarrow n=13$

Thus, there are 13 terms in the given sequence.

(ii) The given A.P. is $-1,\frac{-5}{6}-\frac{2}{3},\frac{1}{2},\dots \dots \frac{10}{3}?$

a = -1

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

$a_n=\frac{10}{3}$

Let there be n terms in the given A.P. then nth term $=\frac{10}{3}$

or $\frac{10}{3}=a_n=a+(n-1)d$

$\Rightarrow \frac{10}{3}=-1+(n-1)(\frac{-5}{6}+1)$

$\Rightarrow \frac{13}{6}=(n-1)\frac{1}{6}$

$\Rightarrow 26=(n-1)$

$\Rightarrow n=27$

Thus, there are 27 terms in the given sequence.