Let a1,a2,a3,… be an arithmetic progression with a1=7 and common difference 8 . Let T1,T2,T3,… be such that T1=3 and Tn+1−Tn=an for n≥1 . Then, which of the following is/are true?

Given T n + 1 − T n = a n n ≥ 1 Put n = 1 , 2 , in succession and adding, we get T 2 − T 1 = a 1 ; T 3 − T 2 = a 2 T n − T n − 1 = a n − 1 So, T n = T 1 + a 1 + a 2 + … + a n − 1 = 3 + 7 + 15 + … to ( n − 1 ) term = 3 + ( n − 1 ) ( 4 n − 1 ) = 4 n 2 − 5 n + 4 , n ≥ 1 Also, ∑ k = 1 n T k = ∑ k = 1 n ( 4 k 2 − 5 k + 4 ) = 4 n ( n + 1 ) ( 2 n + 1 ) 6 − 5 n ( n + 1 ) 2 + 4 n = n ( n + 1 ) 6 { 4 ( 2 n + 1 ) − 15 } + 4 n = n ( n + 1 ) 6 ( 8 n − 11 ) + 4 n Thus, T 20 = 4 × 20 2 − 5 × 20 + 4 = 1504 and T 30 = 4 × 30 2 − 5 × 30 + 4 = 3454 Also we can compute ∑ k = 1 20 T k = 10510 ; ∑ k = 1 30 T k = 35615

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Let $a_{1}, a_{2}, a_{3}, \dots$ be an arithmetic progression with $a_{1} = 7$ and common difference 8 . Let $T_{1}, T_{2}, T_{3}, \dots$ 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?

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

$T_{20} = 1604$

$T_{30} = 3454$

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

answer is B, C.

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Detailed Solution

Given $T_{n + 1} - T_{n} = a_{n} n \geq 1$
Put $n = 1, 2$ , in succession and adding, we get
$T_{2} - T_{1} = a_{1}; T_{3} - T_{2} = a_{2}$
$T_{n} - T_{n - 1} = a_{n - 1}$
So, $T_{n} = T_{1} + a_{1} + a_{2} + \dots + a_{n - 1}$
$= 3 + 7 + 15 + \dots$ to $(n - 1)$ term
$= 3 + (n - 1) (4 n - 1)$
$= 4 n^{2} - 5 n + 4, n \geq 1$
Also,
$\sum_{k = 1}^{n} T_{k} = \sum_{k = 1}^{n} (4 k^{2} - 5 k + 4)$

$= \frac{4 n (n + 1) (2 n + 1)}{6} - 5 \frac{n (n + 1)}{2} + 4 n \begin{array}{l} = \frac{n (n + 1)}{6} {4 (2 n + 1) - 15} + 4 n \\ = \frac{n (n + 1)}{6} (8 n - 11) + 4 n \end{array}$

Thus, $T_{20} = 4 \times 20^{2} - 5 \times 20 + 4 = 1504$
and $T_{30} = 4 \times 30^{2} - 5 \times 30 + 4 = 3454$
Also we can compute
$\sum_{k = 1}^{20} T_{k} = 10510; \sum_{k = 1}^{30} T_{k} = 35615$