The third and fifth term of an arithmetic progression are 10 and-10 respectively.

Question

a)Determine the first and the common difference

t3 = a + 2d = 10
t5 = a + 4d = -10

-a -2d = -10
a + 4d = -10
2d = -20
d = -10
a + 2(-10) = 10
a -20 = 10
a = 30; d = -2
b)The sum of the first 15 terms
I am stuck here for b.

Answer 1

In your last line you have
a = 30, d = -2
but you had found d = -10, which is correct

so a = 30, d = -10

sum(n) = (n/2)(2a + (n-1)d ) , you should know that formula

sum(15) = (15/2)(60 + 14(-10))
= -600

Answer 2

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