The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.

Given:

The 19th term of an A.P. is equal to three times its sixth term.

9th term $=19$

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,

$a_{9}=a+(9-1)d$

$19=a+8d$

$a=19-8d$.....(i)

$a_{19}=a+(19-1)d$

$=a+18d$

$a_{6}=a+(6-1)d$

$=a+5d$

According to the question,

$a_{19}=3\times a_6$

$a+18d=3(a+5d)$

$a+18d=3a+15d$

$3a-a=18d-15d$

$2a=3d$

$2(19-8d)=3d$    (From (i))

$38-16d=3d$

$16d+3d=38$

$19d=38$

$d=2$

This implies,

$a=19-8(2)$

$=19-16$

$=3$

Therefore,

$a_1=3$

$a_2=a+d=3+2=5$

$a_3=a+2d=3+2(2)=3+4=7$

$a_4=a+3d=3+3(2)=3+6=9$

Hence, the required A.P. is $3, 5, 7, 9,......$

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Updated on: 10-Oct-2022

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