# Let a1, a2, a3,.......... a30, be an A.P.

Question:

Let $a_{1}, a_{2}, \ldots \ldots . ., a_{30}$ be an A. P., $S=\sum_{i=1}^{30} a_{i}$ and $\mathrm{T}=\sum_{\mathrm{i}=1}^{15} \mathrm{a}_{(2 \mathrm{i}-1)} .$ If $\mathrm{a}_{5}=27$ and $\mathrm{S}-2 \mathrm{~T}=75$, then

$\mathrm{a}_{10}$ is equal to :

1. 57

2. 47

3. 42

4. 52

Correct Option: , 4

Solution:

$S=a_{1}+a_{2}+\ldots \ldots+a_{30}$

$S=\frac{30}{2}\left[a_{1}+a_{30}\right]$

$S=15\left(a_{1}+a_{30}\right)=15\left(a_{1}+a_{1}+29 d\right)$

$\mathrm{T}=\mathrm{a}_{1}+\mathrm{a}_{3}+\ldots .+\mathrm{a}_{29}$

$=\left(a_{1}\right)+\left(a_{1}+2 d\right) \ldots \ldots+\left(a_{1}+28 \mathrm{~d}\right)$

$=15 \mathrm{a}_{1}+2 \mathrm{~d}(1+2+\ldots .+14)$

$\mathrm{T}=15 \mathrm{a}_{1}+210 \mathrm{~d}$

Now use $\mathrm{S}-2 \mathrm{~T}=75$

$\Rightarrow 15\left(2 \mathrm{a}_{1}+29 \mathrm{~d}\right)-2\left(15 \mathrm{a}_{1}+210 \mathrm{~d}\right)=75$

$\Rightarrow \mathrm{d}=5$

Given $\mathrm{a}_{5}=27=\mathrm{a}_{1}+4 \mathrm{~d} \Rightarrow \mathrm{a}_{1}=7$

Now $\mathrm{a}_{10}=\mathrm{a}_{1}+9 \mathrm{~d}=7+9 \times 5=52$