The sum of the 5th and the 9th terms of an AP is 30. If its 25th term is three times its 8th term, find the AP.

Given:

The sum of the 5th and the 9th terms of an AP is 30.

25th term is three times its 8th term.
To do:

We have to find the AP.

Solution:
Let the first term of the AP be $a$ and the common difference be $d$.

This implies,

nth term $=a+(n-1)d$

5th term$=a+(5-1)d=a+4d$

9th term$=a+(9-1)d=a+8d$

$a+4d+a+8d=30$    (Given)

$2a+12d=30$

$2(a+6d)=2(15)$

$a+6d=15$

$a=15-6d$....(i)

8th term$=a+(8-1)d=a+7d$

25th term$=a+(25-1)d=a+24d$

$a+24d=3(a+7d)$   (Given)

$a+24d=3a+21d$

$3a-a=24d-21d$

$2a=3d$

$2(15-6d)=3d$   (From (i))

$30-12d=3d$

$30=12d+3d$

$15d=30$

$d=\frac{30}{15}$

$d=2$

Therefore,

$a=15-6(2)=15-12=3$

$a+d=3+2=5$

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

The required AP is $3, 5, 7, .....$.

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Updated on: 10-Oct-2022

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