Let $a_{1}, a_{2}, a_{3}, ......$ be an arithmetic progression with $a_{1} = 7$ and common difference 8. Let $T_{1}, T_{2}, T_{3}, ..........$ be such that $T_{1} = 3$ and $T_{n + 1} - T_{n} = a_{n}$ for $n \geq 1, n \in N$ . Then, which of the following is/are TRUE?

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$T_{20} = 1504$

$\sum_{k = 1}^{30} T_{k} = 35610$

$T_{30} = 3454$

$\sum_{k = 1}^{20} T_{k} = 10510$

answer is A, B, C.

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Detailed Solution

$a_{1} = 7, d = 8$
$T_{n + 1} - T_{n} = a_{n} \forall n \geq 1$
$S_{n} = T_{1} + T_{2} + T_{3} + ....... + T_{n - 1} + T_{n}$
$S_{n} = T_{1} + T_{2} + T_{3} + ............ + T_{n - 1} + T_{n}$ on subtraction
$T_{n} = T_{1} + a_{1} + a_{2} + ....... + a_{n - 1}$
$T_{n} = 3 (n - 1) (4 n - 1)$
$T_{n} = 4 n^{2} - 5 n + 4$
$\sum_{k = 1}^{n} T_{k} = 4 Σ n^{2} - 5 Σ n + 4 n$
$T_{20} = 1504$
$T_{30} = 3454$
$\sum_{k = 1}^{30} T_{k} = 35615$

$\sum_{k = 1}^{20} T_{k} = 10510$