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Sum of N terms

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by kmittal82 » Sat Aug 28, 2010 3:26 pm
Simplest way to think about it is this:

Each term of the series ends in 7 or 77, so the number of terms needed to add up to a total of 350 should be a multiple of 10.
(7x10 = 70, 7x20 = 140 etc). As you can see, the only way you can get a bunch of 7s to add up to a 0 in the units place is to add the 7's by a multiple of 10.
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by sirisha.g » Sat Aug 28, 2010 6:31 pm
I did it this way.

if u consider all 7's then the number of terms should be 50(350/7) this is not in the choices.
So, 7*11=77 so replace first 11 terms in 50 terms with one 77 term so u will have 50-11=39+1=40
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by GMATGuruNY » Sat Aug 28, 2010 7:04 pm
JeetGulia wrote:If each term in the sum a1+a1+a3+a4+...+aN is either 7 or 77 and the sum equals 350, which of the following could be equal to n?

1) 38
2) 39
3) 40
4) 41
5) 42

Please help

OA 40
350/7 = 50. So if each term were 7, we'd have 50 terms. The answer choices are all a little less than 50. So most -- but not all -- of the terms will be 7.

If 1 term = 77, we have 350-77 = 273 left.
273/7 = 39. This works!
So 1 term = 77, 39 terms = 7.
Total number of terms is 1+39 = 40.

The correct answer is C.
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by diebeatsthegmat » Sun Aug 29, 2010 10:39 am
JeetGulia wrote:If each term in the sum a1+a1+a3+a4+...+aN is either 7 or 77 and the sum equals 350, which of the following could be equal to n?

1) 38
2) 39
3) 40
4) 41
5) 42

Please help

OA 40
(a1+a2+a3...+a(n-1)+an)*7=350
thus (a1+a2+...+a(n-1)+an=50
suppose a1, a2,....,a(n-1)=1
we will have its sum =39, you know 1+1+1=3 thus a1+a2+a3=3 so the sum from a1 to a39 when all a is 1 is 39
thus we need 11 to get the sum 39+11=50
thus an must be 77 and an must be 40
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