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Coefficients of three consecutive terms in a given ratio: find n and r

This type of question, asked 5 times: 2021 Paper 8, 29 Jan 2023, 8 Apr 2023, 11 Apr 2023, 6 Apr 2024. All 5 questions below, oldest first.

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  1. Question 1

    2021 Paper 8

    86
    Let the coefficients of third, fourth and fifth terms in the expansion of (x+ax2)n,x≠0\left( x+\frac{a}{x^{2}} \right)^{n},x\neq 0, be in the ratio 12:8:312:8:3. Then the term independent of xx in the expansion, is equal to

  2. Question 2

    29 Jan 2023

    84
    Let the coefficients of three consecutive terms in the binomial expansion of (1+2x)n(1 + 2x)^{n} be in the ratio 2:5:82:5:8. Then the coefficient of the term, which is in the middle of these three terms, is

  3. Question 3

    8 Apr 2023

    72
    If the coefficients of the three consecutive terms in the expansion of (1+x)n(1 + x)^{n} are in the ratio 1:5:201:5:20, then the coefficient of the fourth term is

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  4. Question 4

    11 Apr 2023

    166
    The sum of the coefficients of three consecutive terms in the binomial expansion of (1+x)n+2(1 + x)^{n + 2}, which are in the ratio 1:3:51:3:5, is equal to

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  5. Question 5

    6 Apr 2024

    157
    Let 0≤r≤n0 \leq r\leq n. If  n+1Cr+1: nCr: n−1Cr−1=55:35:21\ ^{n+ 1}C_{r+ 1}:\ ^{n}C_{r}:\ ^{n- 1}C_{r- 1}= 55:35:21, then 2n+5r2n+ 5r is equal to:

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