JEE Advanced 2022 — Paper 1 — MATH-II solution

JEE Advanced2022Paper 1MATH-II

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?

A
$T_{20} = 1604$
B
$k = 1 \sum 20 T_{k} = 10510$
C
$T_{30} = 3454$
D
$k = 1 \sum 30 T_{k} = 35610$

Solution

Here $a_{n} = 7 + 8 (n - 1) = 8 n - 1$ . Telescoping $T_{n + 1} - T_{n} = a_{n}$ from $n = 1$ to $n - 1$ :

$T_{n} = T_{1} + k = 1 \sum n - 1 a_{k} = 3 + k = 1 \sum n - 1 (8 k - 1) = 3 + 4 (n - 1) n - (n - 1) = 4 n^{2} - 5 n + 4.$

Equivalently, $T_{n + 1} = 3 + n (4 n + 3)$ , hence $T_{n} = 3 + (n - 1) (4 n - 1) = 4 n^{2} - 5 n + 4$ .

Check (A): $T_{20} = 4 (400) - 100 + 4 = 1504 \neq = 1604$ . So (A) is FALSE.

Check (C): $T_{30} = 4 (900) - 150 + 4 = 3454$ . So (C) is TRUE.

Check (B): $k = 1 \sum 20 T_{k} = k = 1 \sum 20 (4 k^{2} - 5 k + 4) = 4 \cdot \frac{20 \cdot 21 \cdot 41}{6} - 5 \cdot \frac{20 \cdot 21}{2} + 4 \cdot 20 = 11480 - 1050 + 80 = 10510$ . So (B) is TRUE.

Check (D): $k = 1 \sum 30 T_{k} = 4 \cdot \frac{30 \cdot 31 \cdot 61}{6} - 5 \cdot \frac{30 \cdot 31}{2} + 4 \cdot 30 = 37820 - 2325 + 120 = 35615 \neq = 35610$ . So (D) is FALSE.

$∴$ The correct options are (B) and (C).

← Back to full paper

Practice more MATH-II

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →