Question

The first term of A.p is 2 and the common difference is 3 find the sum of n term of the progression. Find the value of n for which the sum of progression is 610. Find also the least value of n for which the sum exceed 1000

Answer 1

so you have a = 2 and d = 3
sum(n) = (n/2)(4 + (n-1)3)
= (n/2)(4 + 3n - 3)
= n(3n + 1)/2

how about
n(3n+1)/2 = 610
n(3n + 1) = 1220
3n^2 + n - 1220 = 0
(n - 20)(3n + 61) = 0
n = 20 or n is negative, which is no good

so the sum of 20 terms is 610

for the last part you need:
n(3n+1)/2 > 1000
3n^2 + n - 2000 > 0
find roots of 3n^2 + n - 2000 = 0
take the next positive whole integer, eg. if the answer is 45.2, use 46

Answer 2

Thanks that just what I needed it's great

Answer 3

My answer is queen