Question

Let $a_1, a_2 , ....... , a_{30}$ be an $A. P$., $S =\displaystyle \sum^{30}_{i=1} a_i $ and $T = \displaystyle\sum^{15}_{i=1} a_{(2i -1)}$. If $a_5 = 27$ and $S - 2T = 75, $ then $a_{10}$ is equal to :

• A. 57
• B. 47
• C. 42
• D. 52

Answer 1

Answer: (D)

$S = a_1 + a_2 + ...... + a_{30}$
$S = \frac{30}{2} [a_1 + a_{30}]$
$S = 15\left(a_{1} +a_{30}\right) = 15 \left(a_{1} +a_{1} +29d\right) $
$ T = a_{1}+a_{3} + ....+ a_{29} $
$ = \left(a_{1}\right)+ \left(a_{1}+2d\right) ....+\left(a_{1} +28d\right) $
$ = 15a_{1} +2d\left(1+2+.....+14\right) $
$T = 15a_{1} +210d $
Now use S - 2T = 75
$ \Rightarrow \; 15 (2a_1 + 29d) - 2 (15a_1 + 210 d) = 75$
$ \Rightarrow \; d = 5$
Given $a_5 = 27 = a_1 + 4d \; \Rightarrow \; a_1 = 7$
Now $a_{10} = a_1 + 9d = 7 + 9 \times 5 = 52$