An important anoriented question from arithmetic progression chapter as it was already asked in previous year paper of 2012 in which we have been asked to find its 22nd term, if 4 times the 4th term of an AP is equal to 18 times its 18th term.
RS Aggarwal, class 10, chapter 5A, question no 32
Let the AP be
a,a+d,a+2d,........,a+(n−1)d
Now, rth term
Tr=a+(r−1)d
∴T4=a+(4−1)d=a+3d
also, T18=a+(18−1)d=a+17d
Given
4(a+3d)=18(a+17d)
⇒4a+12d=18a+306d
⇒14a=−294d
⇒a=−21d
then
⇒T22=a+(n−1)d
=−21d+(22−1)d
=−21d+21d
=0