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# The 5th term of an AP is 22 and its 9th term is six times the 2nd term. Find that AP.

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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