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Terms or sums found from a given formula for the sum of n terms

This type of question, asked 4 times: 12 Apr 2023, 23 Jan 2026 S2, 2 Apr 2026 S1, 5 Apr 2026 S1. All 4 questions below, oldest first.

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  1. Question 1

    12 Apr 2023

    63
    Let ⟨an⟩\left\langle a_{n} \right\rangle be a sequence such that a1+a2+…+an=n2+3n(n+1)(n+2)a_{1}+a_{2}+ \ldots +a_{n}=\frac{n^{2}+ 3n}{(n + 1)(n + 2)}. If 28∑k=1101ak=p1p2p3…pm28\sum_{k = 1}^{10} \frac{1}{a_{k}}=p_{1}p_{2}p_{3}\ldots p_{m}, where p1,p2,…..pmp_{1},p_{2},\ldots..pm are the first mm prime numbers, then mm is equal to

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  2. Question 2

    23 Jan 2026 S2

    70
    Let ∑k=1nak=αn2+βn\sum_{k = 1}^{n} a_{k}= \alpha n^{2}+ \beta n. If a10=59a_{10}= 59 and a6=7a1a_{6}= 7a_{1} then α+β\alpha + \beta is equal to

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  3. Question 3

    2 Apr 2026 S1

    72
    If ∑k=1nak=6n3\sum_{k = 1}^{n} a_{k} = 6n^{3}, then ∑k=16(ak+1−ak36)2\sum_{k = 1}^{6} \left( \frac{a_{k + 1} - a_{k}}{36} \right)^{2} is equal to ____\_\_\_\_ .

  4. Question 4

    5 Apr 2026 S1

    52
    Let the sum of the first nn terms of an A.P. be 3n2+5n3n^{2}+ 5n. Then the sum of squares of the first 10 terms of the A.P. is :

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