JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If $S_n = 4 + 11 + 21 + 34 + 50 + \dots$ to $n$ terms, then $\frac{1}{60} (S_{29} - S_9)$ is equal to:

Choose the correct answer:

Explanation

Step 1: General term ( $a_n$ ) nikalna

Diye gaye terms se:

$n=1: A + B + C = 4$
$n=2: 4A + 2B + C = 11$
$n=3: 9A + 3B + C = 21$

Inhe solve karne par humein milta hai: $A = \frac{3}{2}, B = \frac{5}{2}, C = 0$ .

Isliye:

a_n = \frac{3n^2 + 5n}{2}

Step 2: $S_n$ ka formula

S_n = \sum a_n = \frac{3}{2} \sum n^2 + \frac{5}{2} \sum n

S_n = \frac{3}{2} \left[ \frac{n(n+1)(2n+1)}{6} \right] + \frac{5}{2} \left[ \frac{n(n+1)}{2} \right]

S_n = \frac{n(n+1)(2n+1)}{4} + \frac{5n(n+1)}{4} = \frac{n(n+1)}{4} (2n + 1 + 5)

S_n = \frac{n(n+1)(2n+6)}{4} = \frac{n(n+1)(n+3)}{2}

Step 3: Final Calculation

Humein nikalna hai $\frac{1}{60} (S_{29} - S_9)$ . Note karein ki $S_{29} - S_9 = \sum_{n=10}^{29} a_n$ .

S_{29} = \frac{29 \times 30 \times 32}{2} = 13920

S_9 = \frac{9 \times 10 \times 12}{2} = 540

Ab value put karte hain:

\text{Result} = \frac{1}{60} (13920 - 540)

\text{Result} = \frac{13380}{60} = 223

Choose the correct answer:

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