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 15th term.

Explanation

Let a and be the first term and common difference of an AP, respectively.
Now, by given condition, `a_(8)=(1)/(2)a_(2)`
`implies a+7d=(1)/(2)(a+d) " " [ :' a_(n)=a+(n-1)d]`
`implies 2a+14d=a+d`
` implies a+13d=0 " "` …(i)
and `a_(11)=(1)/(3)a_(4)+1`
`implies a+10d =(1)/(3)[a+3d]+1`
`implies 3a+30d=a+3d+3`
` implies 2a+27d=3 " " `...(ii)
From Eqs. (i) and (ii),
`2(-13d)+27d=3`
` implies -26d +27d=3`
` implies d=3`
From Eq. (i),
`a+13(3)=0`
` implies a= -39`
` :. a_(15)=a+14d= -39+14(3)`
` " "= -39+42=3`