Suneetha P.
asked • 02/21/17How many terms of the A.P.35, 42, 49 …. Are needed to give the sum 500
How many terms of the A.P.35, 42, 49 …. Are needed to give the sum 500
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2 Answers By Expert Tutors
Andrew M. answered • 02/21/17
Tutor
New to Wyzant
Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors
35,42, 49. ....
common difference is d=7
an = a1 + (n-1)d = 35 + 7(n-1) = 35+7n-7 = 28+7n
Sum of an arithmetic sequence:
n
∑ai = (n/2)(a1 + an)
i=1
(n/2)(35 + 28+7n) = 500
n(63+7n) = 1000
7n2 + 63n - 1000 = 0
n = [-63 ±√(632 - 4(7)(-1000))]/2(7)
n = (-63 ±√31969)/14
n ≅ 8.3
The trouble with your sequence is that the sum of
this sequence will never be exactly 500
The sum of the 1st 8 terms is
35+42+49+56+63+70+77+84 = 476
The sum of the 1st 9 terms is 476+91 = 567
That is why n came out to a decimal
instead of an integer
Kenneth S. answered • 02/21/17
Tutor
4.8
(62)
Expert Help in Algebra/Trig/(Pre)calculus to Guarantee Success in 2018
Sn = ½•n[2a1 + (n-1)d]
500 = (n/2)[2•35 + (n-1)(7)]
This is quadratic equation involving n, # of terms. Solve it by your preferred method.
Andrew M.
Just to show that this is the same quadratic
that I came up with in my solution:
500 = (n/2)[2*35 + (n-1)(7)]
1000 = n(70 + 7n - 7)
1000 = n(7n + 63)
7n2 + 63n - 1000 = 0
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02/21/17
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Michael J.
02/21/17