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1)AP has U6=12.5 and U19=32
a+5d=12.5 (&)
a+18d=32 (2)
(2)-(1), d=1.5
therefore, a=5
(a)What is the nth term?Tn=5+(n-1)1.5
(b)What is the sum of 4 terms between U10 and U13 inclusive?

(a)What is the nature of this sequence?
(b)Calculate the sum of the first nine terms?

(a)Solve SUMn=0
(b)If Wm=2+5m, when is Wm first a three digit number?
(c)Find m for the case where Wm=U10

  • Math -

    for 1. a) I would simplify a bit further.
    = 5 + 1.5 n - 1.5
    = 1.5n + 3.5

    for b)

    find the sum of the first 13 terms, then
    find the sum of the first 9 terms.

    2. Clearly a geometric sequence if the expression is (1/2)^(n-3)
    The problem is that you did not specify the starting value of n
    e.g. if you start with n = 4, term 1 is 1/2
    if you start with n = 1 then the first term is 4
    In any case r = 1/2
    Use you sum formula to find the sum of 9 terms.

    3. obviously an arithmetic sequence with a= 11.5 , d= -.5

    Sum(n) = n/2[2a + (n-1)d]
    0 = n/2[23 - .5(n-1)]

    solve for n

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