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JEE Mathematics 2023 Question with Solution

If Sn=4+11+21+34+50+…S_n = 4 + 11 + 21 + 34 + 50 + \dots to nn terms, then 160(S29−S9)\frac{1}{60}(S_{29} - S_9) is equal to:

  • A

    220220

  • B

    227227

  • C

    226226

  • D

    223223

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The sequence is 4,11,21,34,50,…4, 11, 21, 34, 50, \dots and SnS_n is the sum to nn terms.

Find: 160(S29−S9)\frac{1}{60}(S_{29} - S_9).

First find the general term. The first differences are 7,10,13,16,…7, 10, 13, 16, \dots and the second differences are constant: 3,3,3,…3, 3, 3, \dots. Therefore, the term is quadratic, so let

Tn=an2+bn+cT_n = an^2 + bn + c

Using the first three terms:

T1=a+b+c=4T_1 = a + b + c = 4 T2=4a+2b+c=11T_2 = 4a + 2b + c = 11 T3=9a+3b+c=21T_3 = 9a + 3b + c = 21

Solving, we get

a=32,b=52,c=0a = \frac{3}{2}, \quad b = \frac{5}{2}, \quad c = 0

Hence,

Tn=3n2+5n2T_n = \frac{3n^2 + 5n}{2}

Now sum the terms:

Sn=∑k=1nTk=12∑k=1n(3k2+5k)S_n = \sum_{k=1}^{n} T_k = \frac{1}{2}\sum_{k=1}^{n}(3k^2 + 5k) Sn=12(3∑k=1nk2+5∑k=1nk)S_n = \frac{1}{2}\left(3\sum_{k=1}^{n} k^2 + 5\sum_{k=1}^{n} k\right)

Using

∑k=1nk2=n(n+1)(2n+1)6,∑k=1nk=n(n+1)2\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{k=1}^{n} k = \frac{n(n+1)}{2}

we get

Sn=12(3⋅n(n+1)(2n+1)6+5⋅n(n+1)2)S_n = \frac{1}{2}\left(3\cdot \frac{n(n+1)(2n+1)}{6} + 5\cdot \frac{n(n+1)}{2}\right) Sn=n(n+1)(n+3)2S_n = \frac{n(n+1)(n+3)}{2}

Now evaluate:

S29=29⋅30⋅322=13920S_{29} = \frac{29 \cdot 30 \cdot 32}{2} = 13920 S9=9⋅10⋅122=540S_9 = \frac{9 \cdot 10 \cdot 12}{2} = 540

Therefore,

S29−S9=13920−540=13380S_{29} - S_9 = 13920 - 540 = 13380 160(S29−S9)=1338060=223\frac{1}{60}(S_{29} - S_9) = \frac{13380}{60} = 223

Therefore, the correct option is D and the value is 223223.

Quick Observation

Given: The sequence has constant second difference.

Find: 160(S29−S9)\frac{1}{60}(S_{29} - S_9).

Since the second difference is constant, assume a quadratic term directly:

Tn=an2+bn+cT_n = an^2 + bn + c

Find a,b,ca, b, c from the first three terms, then use standard formulas for ∑k\sum k and ∑k2\sum k^2. This is faster than writing many terms manually because polynomial sequences reduce neatly under summation.

After simplification,

Tn=3n2+5n2T_n = \frac{3n^2 + 5n}{2}

and hence

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

Substitute n=29n=29 and n=9n=9 to get the result.

This gives 223223, so the correct option is D.

Common mistakes

  • Assuming the sequence is arithmetic. The first differences are not constant, so an arithmetic progression formula does not apply. Check second differences before deciding the form of the general term.

  • Finding TnT_n correctly but forgetting that SnS_n is the sum of terms, not the term itself. After getting TnT_n, you must still sum from k=1k=1 to nn.

  • Using incorrect summation formulas for ∑k\sum k or ∑k2\sum k^2. This gives a wrong expression for SnS_n. Use ∑k=1nk=n(n+1)2\sum_{k=1}^{n} k = \frac{n(n+1)}{2} and ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}.

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