Search any question & find its solution
Question:
Answered & Verified by Expert
If 9 times the $9^{\text {th }}$ term of an AP is equal to 13 times the $13^{\text {th }}$ term, then the $22^{\text {nd }}$ term of the AP is
Options:
Solution:
1837 Upvotes
Verified Answer
The correct answer is:
0
0
Suppose the first term is a and common difference is
$d$
$$
\begin{aligned}
&\text { As, } 9 T_9=13 T_{13} \Rightarrow 9(a+8 d)=13(a+12 d) \\
&\Rightarrow(9 a-13 a)=156 d-72 d \Rightarrow-4 a=84 d \\
&\Rightarrow a=-21 d
\end{aligned}
$$
So, $a+21 d=0$
Therefore, $22^{\text {nd }}$ term of the $\left.\mathrm{AP}, \mathrm{T}_{22}=\mathrm{a}+21 d\right]$ $\mathrm{T}_{22}=0$
$d$
$$
\begin{aligned}
&\text { As, } 9 T_9=13 T_{13} \Rightarrow 9(a+8 d)=13(a+12 d) \\
&\Rightarrow(9 a-13 a)=156 d-72 d \Rightarrow-4 a=84 d \\
&\Rightarrow a=-21 d
\end{aligned}
$$
So, $a+21 d=0$
Therefore, $22^{\text {nd }}$ term of the $\left.\mathrm{AP}, \mathrm{T}_{22}=\mathrm{a}+21 d\right]$ $\mathrm{T}_{22}=0$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.