the sum of the first n terms of an arithmetic series is 2n(10-n)

Question

a.) what is the first term of the series
b.) what is the common difference of the series
given that Sn > -50
c.) i.) write down an ineualityt satisfied by n
ii.) hence find the largest value of n for which Sn > -50

Answer 1

The formula for the sum is
sum(n) = (n/2)(2a + (n-1)d (
we have:
2n(10-n) = (n/2)(2a + (n-1)d)
divide by n and expand
20 - 2n = (1/2)(2a + nd - d)
20 - 2n = a + nd/2 - d/2
times 2
40 - 4n = 2a + nd - d
-6a = nd - d - 40
a = (40 - nd + d)/6

b) from -6a = nd - d - 40
-6a = d(n-1) - 40
d = (40-6a)/(n-1)

c) we are told sum(n) > -50
2n(10-n) > -50
n(10-n) > -25
10n - n^2 + 25 > 0
n^2 - 10n - 25 < 0
-2.07 < n < 12.07 , but n is a whole number,
so n < 12

The largest number of terms is 12