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Question:
If 9 times the $9^{\text {th }}$ term of an AP is equal to 13 times the $13^{\text {th }}$ term, show that its $22^{\text {nd }}$ term is 0 .
Solution:
Given : $9 \times\left(9^{\text {th }}\right.$ term $)=13 \times\left(13^{\text {th }}\right.$ term $)$
To prove: $22^{\text {nd }}$ term is 0
$9 \times(a+8 d)=13 \times(a+12 d)$
$9 a+72 d=13 a+156 d$
$-4 a=84 d$
$a=-21 d \ldots . .$ Equation 1
Also $22^{\text {nd }}$ term is given by
a + 21d
Using equation 1 we get
$-21 d+21 d=0$
Hence proved 22nd term is 0.
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