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by Imsukhi » Tue Jul 09, 2013 7:42 am
Find the least value of n for which 3+6+9+...+ to n terms exceeds 500 .

a-18
b-20
c-21
d-22
e-none of above

pls explain
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by sparkles3144 » Tue Jul 09, 2013 9:42 am
Is the answer A?

Sum of n numbers

Sn = (n/2) (2a1 + (n-1)d)

d = difference between terms = 6-3 = 3
n = number of terms
a1 = first term = 3

Sn = sum > 500

Sn = (n/2) (2a1 + (n-1)d)
500 > (n/2) (2a1 + (n-1)d)
500 > (n/2) (2 X 3 + (n-1)3)
500 > (n/2) (6 + 3n - 3)
500 > (n/2) (3 + 3n)

Substitue 17 and 18

When you substitue 17 (number of terms), the sum will be less than 500

When you substitue 18 (number of terms), the sum will be greater than 500
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by Matt@VeritasPrep » Tue Jul 09, 2013 3:57 pm
Let's simplify the question first.

The question is essentially asking for the smallest value of n for which the following equation is true:

3*1 + 3*2 + 3*3 + ... + 3*n > 500

The sum of any series is (Average of terms) * (Number of terms). The average of an arithmetic progression (like this one) is (Greatest Term + Least Term)/2, and the number of terms in this series is n, so the sum is

(3n + 3)
-------- * n
2

Simplifying, we get

3n^2 + 3n > 1000

Or n^2 + n > 1000/3

Or n * (n+1) > 1000/3

1000/3 = 333.33...

Hence n * (n+1) must be at least 334. The smallest n that works is n = 18, as 18 * 19 = 342.

The answer is A
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