(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 $0.$