In another case, p and 2p are the first and second term respectively of an arithmetic progression. The nth term is 336 and the of the first

Question

In another case, p and 2p are the first and second term respectively of an arithmetic progression. The nth term is 336 and the of the first n terms is 7224. Write down two equations in n and p and hence find the values of n and p​

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Jezebel 3 years 2021-07-20T05:18:08+00:00 1 Answers 60 views 0

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    2021-07-20T05:19:20+00:00

    Consecutive terms in this sequence differ by p.

    First term: p

    Second term: p + p = 2p

    Third term: 2p + p = 3p

    and so on. It follows that the n-th term satisfies

    np = 336

    Presumably you meant to say the “sum of the first n terms” is 7224, which is to say

    p + 2p + 3p + … + np = 7224

    which can be rewritten as

    p (1 + 2 + 3 + … + n) = 7224

    p (n (n + 1)/2) = 7224

    n (n + 1) p = 14,448

    Substitute the first equation in the second one and solve for n :

    336 (n + 1) = 14,448

    n + 1 = 43

    n = 42

    Now solve for p :

    42p = 336

    p = 8

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