The 8th term of an $ \mathrm{AP} $ is 31 and its 15th term exceeds its 11th term by 16. Find that AP.

Given:

The 8th term of an \( \mathrm{AP} \) is 31 and its 15th term exceeds its 11th term by 16.

To do:

We have to find the AP.

Solution:

Let the first term of the A.P. be $a$ and the common difference be $d$.

We know that,

nth term of an A.P. $a_n=a+(n-1)d$

Therefore,

$a_{8}=a+(8-1)d$

$31=a+7d$......(i)

$a_{15}=a+(15-1)d$

$=a+14d$.......(ii)

$a_{11}=a+(11-1)d$

$=a+10d$......(iii)

According to the question,

$a_{15}=a_{11}+16$

$a+14d=(a+10d)+16$

$14d+a-a-10d=16$

$4d=16$

$d=4$.....(iv)

$\Rightarrow 31=a+7d$      (From (i))

$31=a+7(4)$

$a=31-28$    

$a=3$

$a_2=a+d=3+4=7$

$a_3=a_2+d=7+4=11$

The required AP is $3, 7, 11, ...........$.

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Updated on: 10-Oct-2022

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