A)
a=5, d=5 , n = ? , Sum(n) ≥ 400 000
(n/2)(10 + (n-1)(5)) ≥ 400 000
n(5n + 5) ≥ 800 000
5n^2 + 5n - 800 000 ≥ 0
n^2 + n - 160 000 ≥ 0
Now I took √160000 = 400
so if the above had been
n^2 + n - 159999 it would have factored to
(n+401)(n-399)
so our equation must have a solution between 399 and 401.
Since n must be a whole number,
sum(399) = (399/2)(10 + 398(5)) = 399000
sum (400) = (400/2)(10 + 399(5)) = 399200
sum(401) = (401/2)(10 + 400(5)) = 403005
So you need 401 terms to exceed 400000
B) "the sum of the first five terms is 30" ---> (5/2)(2a + 4d) = 30
2a + 4d = 12
a + 2d = 6
"the third term is equal to the sum of the first two" ---> a+2d = a + a+d
a = d
sub that into first equation:
d + 2d = 6
d = 2 , then a = 2
first 5 terms: 2 , 4 , 6, 8, and 10
A) which down the first G.P 5+10+20+.....to exceed 400000?
B) in an A.P,the sum of the first five term is 30, and the third term is equal to the sum of the first two. Write the first five terms of the progression.
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