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 1 . Then, which of the following is/are TRUE? [2022]

Question

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? [2022]

Smart Achievers · Accepted Answer

(2, 3)

Given, $a_{1} = 7, d = 8$

$Hence, a_{n} = 7 + (n - 1) 8 and T_{1} = 3$

$Also T_{n + 1} = T_{n} + a_{n}$

$T_{n} = T_{n - 1} + a_{n - 1}$

$⋮$

$T_{2} = T_{1} + a_{1}$

$∴$ $T_{n + 1} = (T_{n - 1} + a_{n - 1}) + a_{n}$

$= T_{n - 2} + a_{n - 2} + a_{n - 1} + a_{n}$

$⋮$

$\Rightarrow T_{n + 1} = T_{1} + a_{1} + a_{2} + \dots + a_{n}$

$\Rightarrow T_{n + 1} = T_{1} + \frac{n}{2} [2 (7) + (n - 1) 8]$

$\Rightarrow T_{n + 1} = 3 + n (4 n + 3) ...(i)$

$Hence, for n = 19; T_{20} = 3 + (19) (79) = 1504$

$For n = 29; T_{30} = 3 + (29) (119) = 3454 \to (c)$

$\sum_{k = 1}^{20} T_{k} = 3 + \sum_{k = 2}^{20} T_{k} = 3 + \sum_{k = 1}^{19} (3 + 4 n^{2} + 3 n)$

$= 3 + 3 (19) + \frac{3 (19) (20)}{2} + \frac{4 (19) (20) (39)}{6}$

$= 3 + 10507 = 10510 \to (b)$

$And similarly$ $\sum_{k = 1}^{30} T_{k} = 3 + \sum_{k = 1}^{29} (4 n^{2} + 3 n + 3) = 35615$

Solution

Q.
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 [2022]

1 $T_{20} = 1604$

2 $\sum_{k = 1}^{20} T_{k} = 10510$

3 $T_{30} = 3454$

4 $\sum_{k = 1}^{30} T_{k} = 35610$

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Solution

Q. 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 [2022]

1 T20=1604

2 ∑k=120Tk=10510

3 T30=3454