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The 9th term of an A.P. is equal to 6 times its second term. If its 5th term is 22, find the A.P.
Given:
The 9th term of an A.P. is equal to 6 times its second term.
5th term $=22$
To do:
We have to find the A.P.
Solution:
Let the required A.P. be $a, a+d, a+2d, ......$
Here,
$a_1=a, a_2=a+d$ and Common difference $=a_2-a_1=a+d-a=d$
We know that,
$a_n=a+(n-1)d$
Therefore,
$a_{5}=a+(5-1)d$
$22=a+4d$
$a=22-4d$.....(i)
$a_{9}=a+(9-1)d$
$=a+8d$
$a_{2}=a+(2-1)d$
$=a+d$
According to the question,
$a_{9}=6\times a_2$
$a+8d=6(a+d)$
$a+8d=6a+6d$
$6a-a=8d-6d$
$5a=2d$
$5(22-4d)=2d$ (From (i))
$110-20d=2d$
$20d+2d=110$
$22d=110$
$d=\frac{110}{22}=5$
This implies,
$a=22-4(5)$
$=22-20$
$=2$
Therefore,
$a_1=2$
$a_2=a+d=2+5=7$
$a_3=a+2d=2+2(5)=2+10=12$
$a_4=a+3d=2+3(5)=2+15=17$
Hence, the required A.P. is $2, 7, 12, 17,......$
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