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 ?

(B) $\sum_{k=1}^{20} T_{k}=10510$; (C) $T_{30}=3454$

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

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Sequences and Series

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 ?

Answer: (B), (C)

Given

a_{n} = 7 + (n - 1) \cdot 8 = 8 n - 1

and

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

.

Summing from

n = 1

n - 1

T_{n} = T_{1} + i = 1 \sum n - 1 a_{i} = 3 + i = 1 \sum n - 1 (8 i - 1)

= 3 + 8 \cdot \frac{( n - 1 ) n}{2} - (n - 1) = 3 + 4 n (n - 1) - (n - 1)

= 3 + (n - 1) (4 n - 1) = 4 n^{2} - 5 n + 4

Checking the options:

(A) $T_{20} = 4 (400) - 100 + 4 = 1504$ ✗
(B) $\sum_{k = 1}^{20} T_{k} = 4 \sum k^{2} - 5 \sum k + 80 = 4 \cdot \frac{20 \cdot 21 \cdot 41}{6} - 5 \cdot \frac{20 \cdot 21}{2} + 80 = 11480 - 1050 + 80 = 10510$ ✓
(C) $T_{30} = 4 (900) - 150 + 4 = 3454$ ✓
(D) $\sum_{k = 1}^{30} T_{k} = 4 \cdot \frac{30 \cdot 31 \cdot 61}{6} - 5 \cdot \frac{30 \cdot 31}{2} + 120 = 37820 - 2325 + 120 = 35615$ ✗

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · One or more correct options