JEE Advance - Mathematics (2022 - Paper 1 Online - No. 10)
Let $$a_{1}, a_{2}, a_{3}, \ldots$$ be an arithmetic progression with $$a_{1}=7$$ and common difference 8. Let $$T_{1}, T_{2}, T_{3}, \ldots$$ be such that $$T_{1}=3$$ and $$T_{n+1}-T_{n}=a_{n}$$ for $$n \geq 1$$. Then, which of the following is/are TRUE ?
$$T_{20}=1604$$
$$\sum\limits_{k=1}^{20} T_{k}=10510$$
$$T_{30}=3454$$
$$\sum\limits_{k=1}^{30} T_{k}=35610$$
Explanation
Here $a_n=7+(n-1) 8$ and $\mathrm{T}_1=3$, $a_1=7, d=8$
Also, $ \mathrm{T}_{n+1}=\mathrm{T}_n+a_n$
$\mathrm{T}_n=\mathrm{T}_{n-1}+a_{n-1}$
$\mathrm{T}_2=\mathrm{T}_1+a_1$
$\therefore \mathrm{T}_{n+1}=\left(\mathrm{T}_{n-1}+a_{n-1}\right)+a_n$
$\mathrm{T}_{n+1}=\mathrm{T}_{n-2}+a_{n-2}+a_{n-1}+a_n$
So
$$ \begin{aligned} & \mathrm{T}_{n+1}=\mathrm{T}_1+a_1+a_2 \ldots \ldots a_n \\\\ & \mathrm{~T}_{n+1}=\mathrm{T}_1+\frac{n}{2}(2 \times 7+(n-1) 8) \\\\ & \mathrm{T}_{n+1}=\mathrm{T}_1+n(4 n+3) ........(1) \end{aligned} $$
For (A), if $ n=19, \mathrm{~T}_{20}=3+(19)(79)=1504$
For (C), if $n=29, \mathrm{~T}_{30}=3+29(119)=3454$
For (B),
$$ \begin{aligned} & \sum_{k=1}^{20} \mathrm{~T}_k =\sum_{k=1}^{20}\left(\mathrm{~T}_1+4 n^2+3 n\right)+3 \\\\ & = 3+\sum_{k=1}^{20}\left(3+4 n^2+3 n\right) \\\\ &= 3+3(19)+\frac{3(19)(20)}{2} +\frac{4(19)(20)(34)}{6} \\\\ &= 3+10507=10510 \end{aligned} $$
Similarly, for (D)
$$ \sum\limits_{k=1}^{30} \mathrm{~T}_k=3+\sum\limits_{k=1}^{29}\left(4 n^2+3 n+3\right)=35615 $$
Also, $ \mathrm{T}_{n+1}=\mathrm{T}_n+a_n$
$\mathrm{T}_n=\mathrm{T}_{n-1}+a_{n-1}$
$\mathrm{T}_2=\mathrm{T}_1+a_1$
$\therefore \mathrm{T}_{n+1}=\left(\mathrm{T}_{n-1}+a_{n-1}\right)+a_n$
$\mathrm{T}_{n+1}=\mathrm{T}_{n-2}+a_{n-2}+a_{n-1}+a_n$
So
$$ \begin{aligned} & \mathrm{T}_{n+1}=\mathrm{T}_1+a_1+a_2 \ldots \ldots a_n \\\\ & \mathrm{~T}_{n+1}=\mathrm{T}_1+\frac{n}{2}(2 \times 7+(n-1) 8) \\\\ & \mathrm{T}_{n+1}=\mathrm{T}_1+n(4 n+3) ........(1) \end{aligned} $$
For (A), if $ n=19, \mathrm{~T}_{20}=3+(19)(79)=1504$
For (C), if $n=29, \mathrm{~T}_{30}=3+29(119)=3454$
For (B),
$$ \begin{aligned} & \sum_{k=1}^{20} \mathrm{~T}_k =\sum_{k=1}^{20}\left(\mathrm{~T}_1+4 n^2+3 n\right)+3 \\\\ & = 3+\sum_{k=1}^{20}\left(3+4 n^2+3 n\right) \\\\ &= 3+3(19)+\frac{3(19)(20)}{2} +\frac{4(19)(20)(34)}{6} \\\\ &= 3+10507=10510 \end{aligned} $$
Similarly, for (D)
$$ \sum\limits_{k=1}^{30} \mathrm{~T}_k=3+\sum\limits_{k=1}^{29}\left(4 n^2+3 n+3\right)=35615 $$
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