Question

Consider the sequence $S=7+13+21+31+\cdots +T_n$. The value of $T_{70}$ is

• A. $5013$
• B. $5050$
• C. $5113$
• D. $5213$

Answer 1

Answer: (C)

$5113$

Given $S=7+13+21+31+....+T_n$.
$\Rightarrow S=7+13+21+31+....+T_{n-1}+T_n$
$\Rightarrow 0=(7+6+8+10+....+n^{th}term)-T_n$
$\Rightarrow T_n=(7+6+8+10+....+n^{th}term)$
$\Rightarrow T_n=3+(4+6+8+10+....+n^{th}term)$
Since the series $(4+6+8+10+....+n^{th}term)$ is A.P.
With the first term $a=4$ and the common difference $d=2$.
Therefore, $T_n=3+\frac{n}{2}\left[2\right(4)+(n-1\left)2\right]$
$\Rightarrow T_n=3+\frac{n}{2}[8+(n-1\left)2\right]$
$\Rightarrow T_n=3+\frac{n}{2}[2n+6]$
$\Rightarrow T_n=3+n(n+3)$
$\Rightarrow T_n=n^2+3n+3$
Now, put $n=70$; we get
$T_{70}={\left(70\right)}^2+3\left(70\right)+3$
$\Rightarrow T_{70}=4900+210+3$
$\Rightarrow T_{70}=5113$