Question

The eight term of an A.P. is half of its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.

Answer 1

Given $T_8$ = $\frac{1}{2}$ $T_2$

$\Rightarrow$ a + 7d = $\frac{a+d}{2}$

$\Rightarrow$ 2(a+7d) =a+d
$\Rightarrow$ 2a + 14d = a + d
$\Rightarrow$ a = -13d

Again given $T_{11}$ = $\frac{1}{3}$ $T_4$ + 1
$\Rightarrow$ a +10d = $\frac{a+3d}{3}$ + 1
$\Rightarrow$ a +10d -1 = $\frac{a+3d}{3}$
$\Rightarrow$ 3(a + 10d -1) = a +3d
$\Rightarrow$ 3a + 30d -3 = a+ 3d
$\Rightarrow$ 2a + 27d =3

$\Rightarrow$ 2(-13d) + 27d =3 ............ (since a = -13d)
$\Rightarrow$ -26d + 27d = 3
$\Rightarrow$ d = 3
$\Rightarrow$ a= -13d ..............(since d=3)
$\Rightarrow$ a = -39

Now $T_{15}$ = a + 14d

$\Rightarrow$ = -39 + 14x3
$\Rightarrow$ = -39 + 42
$\Rightarrow$ hence$T_{15}$ =3