(ii) Which term of the AP $84,80,76,$ … is $0?$

An arithmetic progression or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant.

Solution:

(i) Given A.P. is $3,8,13,$ ………..

First term $\left( a \right)=3$

Common difference $\left( d \right)=$ Second term – first term

$=8-3$

$=5$

We know that, ${{n}^{th}}$ term $\left( {{a}_{n}} \right)=a+\left( n-1 \right)d$

And, given ${{n}^{th}}$ term ${{a}_{n}}=248$

$248=3+\left( n-1 \right)5$

$248=-2+5n$

$5n=250$

$n=250/5=50$

$\therefore {{50}^{th}}$ term in the A.P is $248.$

(ii) Given A. P is $84,80,76,$ …………

First term $\left( a \right)=84$

Common difference $\left( d \right)={{a}_{2}}-a$

$=80-84$

$=-4$

We know that, ${{n}^{th}}$ term $\left( {{a}_{n}} \right)=a+\left( n-1 \right)d$

And, given nth term is $0$

$0=84+\left( n-1 \right)-4$

$84=+4\left( n-1 \right)$

$n-1=84/4=21$

$n=21+1=22$

$\therefore {{22}^{nd}}$ term in the A.P is $0.$