if sum of first n terms of an A.P.is 5(n)^2+3(n) find the A.P.& n^th term.?

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  • Puggy
    Lv 7
    1 decade ago
    Favorite Answer

    So it is given that

    S(n) = 5n^2 + 3n

    And you want to find the arithmetic progression and the nth term.

    Recall that the sum of an arithmetic progression is given by the formula:

    S(n) = n(2[a1] + (n - 1)d) / 2

    {Note: a1 represents the unknown first term, and d represents the difference between consecutive terms}

    Through algebraic manipulation with which I won't show the details, we can replace S(n) to be

    S(n) = (d/2)n^2 + n(a1 - d/2)

    BUT

    S(n) = 5n^2 + 3n, so all we have to do is equate the coefficients of n^2 and the coefficients of n.

    d/2 = 5 (implies d = 10)

    a1 - d/2 = 3 (implies a1 = 8)

    Therefore, a1 (our first term) is equal to 8, and d = 10.

    Our sequence then goes as follows;

    8, 18, 28, 38, 48, ...

    with our general term being 8 + (n - 1)10, or (8 + 10n - 10), or

    (10n - 2)

    Conclusion:

    The arithmetic progression is the sequence with a1 = 8 and r = 10, and

    a[n] = (10n - 2)

    Edit:

    ---

    If you're interested in knowing the details of how I got from

    S(n) = n(2[a1] + (n - 1)d) / 2

    to

    S(n) = (d/2)n^2 + n(a1 - d/2)

    Here it is, below.

    S(n) = n(2[a1] + (n - 1)d) / 2

    Factor out a 1/2, since it's a fraction anyway.

    S(n) = (n/2) (2[a1] + (n - 1)d)

    Expand the inside.

    S(n) = (n/2) (2[a1] + nd - d)

    Distribute the (n/2)

    S(n) = (a1)n + d(n^2)/2 - dn/2

    Group together the n^2 terms and the n terms.

    S(n) = d(n^2)/2 + (a1)n - dn/2

    S(n) = (d/2)n^2 + (a1 - d/2)n

    It was only through this algebraic manipulation that we could equate this component-wise with our given sum.

  • 1 decade ago

    Sn = 5 n^2 + 3n

    S1 = 5 + 3

    = 8 is the first term(a)

    S2 = 5*4 + 3*2

    = 26 is the sum of first two terms

    second term = 26 -8

    = 18

    S3 = 5*9 + 9

    = 54 is the sum of first three terms

    third term = 54 - 26

    = 28

    A.P is 8,18,28,..

    a= 8, d = 10

    nth term = a + (n-1)d

    = 8 + (n-1)10

    = 10n -2

  • 1 decade ago

    becuase it is ap let kth term be ak+b

    sum of n terms a n(n+1)/2 + bn = an^2/2 + (b+a/2) n

    comparing with 5n^2 + 3n

    a/2 = 5 oe a = 10 and b+5 = 3 or b = -2

    nth term = 10n - 2

  • Wal C
    Lv 6
    1 decade ago

    S_n = 5n² + 3n

    S_(n - 1) = 5(n - 1)² + 3(n - 1)

    = 5n² - 10n + 5 + 3n - 3

    = 5n² - 7n + 2

    Now S_n = S_(n - 1) + U_n

    So U_n = S_n - S_(n - 1)

    = 10n - 2

    So the AP is {8, 18, 28, ...., 10n - 2, ....}

    Source(s): Me ;^))
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  • Helmut
    Lv 7
    1 decade ago

    Xn = 5(n)^2 + 3(n) - 5(n-1)^2 - 3n + 3

    Xn = 5(n)^2 + 3(n) - 5n^2 + 10n - 5 - 3n + 3

    Xn = 10n - 2

    N = ∑ 8 + 18 + 28 + . . . . . + (10n - 2)

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