The eighth term of an A.P is half its second term and the eleventh term exceeds one third of its fourth term by $1$. Find the $15^{th}$ term of the A.P.
Let the first term of the A.P be $\;$ $t_1 = a$ $\;$ and the common difference $= d$
$n^{th}$ $\;$ term of A.P $\;$ $= t_n = a + \left(n - 1\right) d$
$\therefore \;$ $2^{nd}$ term of A.P $= t_2 = a + d$;
$4^{th}$ term $= t_4 = a + 3d$;
$8^{th}$ term $= t_8 = a + 7d$; $\;\;$ and
$11^{th}$ term $= t_{11} = a + 10d$
As per the question,
$t_8 = \dfrac{1}{2} \times t_2$
i.e. $\;$ $a + 7d = \dfrac{1}{2} \times \left(a + d\right)$
i.e. $\;$ $2a + 14 d = a + d$ $\;\;\;$
i.e. $\;$ $a + 13d = 0$ $\implies$ $a = -13d$ $\;\;\; \cdots \;\;\; (1)$
Also, as per the problem
$t_{11} = \left(\dfrac{1}{3} \times t_4\right) + 1$
i.e. $\;$ $a + 10 d = \dfrac{1}{3} \times \left(a + 3d\right) + 1$
i.e. $\;$ $3a + 30d = a + 3d + 3$
i.e. $\;$ $2a + 27d = 3$ $\;\;\; \cdots \;\;\; (2)$
In view of equation $(1)$, equation $(2)$ becomes
$2 \times \left(-13 d\right) + 27 d = 3$ $\;\;\;$ $\implies$ $d = 3$
Substituting the value of $d$ in equation $(1)$ gives
$a = -13 \times 3 = -39$
Now, $15^{th}$ term of A.P $= t_{15} = a + 14d$
i.e. $\;$ $t_{15} = -39 + \left(14 \times 3\right) = 3$