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# The eighth term of an AP is half its second term and the eleventh term exceeds one third of its fourth term by 1 . Find the $ 15^{\text {th }} $ term.

Given:

The eighth term of an A.P. is half of its second term and the eleventh term exceeds one third of its fourth term by 1.

To do:

We have to find the 15th term.

Solution:

Let the first term, common difference and the number of terms of the given A.P. be $a, d$ and $n$ respectively.

We know that,

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

Therefore,

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

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

$a_{2}=a+(2-1)d$

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

According to the question,

$a_{8}=\frac{1}{2}a_2$

$a+7d=\frac{1}{2}(a+d)$

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

$2a+14d-a-d=0$

$a+13d=0$

$a=-13d$....(iii)

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

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

$a_{4}=a+(4-1)d$

$=a+3d$....(v)

According to the question,

$a_{11}=\frac{1}{3}a_4+1$

$a+10d=\frac{1}{3}(a+3d)+1$

$a+10d=\frac{a+3d+1\times3}{3}$

$3(a+10d)=a+3d+3$

$3a+30d-a-3d-3=0$

$2a+27d-3=0$

$2(-13d)+27d-3=0$ (From (iii))

$-26d+27d-3=0$

$d=3$

Substituting $d=3$ in (iii), we get,

$a=-13(3)$

$a=-39$

15th term $a_15=a+(15-1)d$

$=-39+14(3)$

$=-39+42$

$=3$

The 15th term is $3$.