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The sum of 8 consecutive integers is 180. [#permalink]
Bunuel VeritasKarishma chetan2u GMATPrepNow

I respect all above solutions, but in my view I still feel they are prone to calculation errors.
Any way I can use below facts to solve this problem:
1. For consecutive integers: average = (First term + last term) / 2
2. Even for using simple average equation (avg = sum / quantity) ; since 180 / 8 gives me a decimal,
I do not feel like proceeding since I make calculations mistakes in same.

May I know your approach please?
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The sum of 8 consecutive integers is 180. [#permalink]
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adkikani wrote:
The sum of 8 consecutive integers is 180. What is the sum of first 4 of these integers?

a. 80
b. 82
c. 85
d. 86
e. 90


Here, all the methods are almost equivalent in the amount of time and effort required.

\(\frac{(a + a + 7)}{2} = \frac{180}{8}\)

2a = 45 - 7 = 38

a = 19

Sum of first four numbers = 4a + 1 + 2 + 3 = 4*19 + 6 = 82

Answer (B)

Another method:

Each of the last 4 numbers will be 4 more than the previous 4 numbers hence the sum will have 16 extra from the last 4 numbers.
180 - 16 = 164 is the sum of all 8 numbers if the last 4 numbers were the same as first 4 numbers.
So sum of first 4 numbers = 164/2 = 82

Originally posted by KarishmaB on 23 Aug 2019, 05:00.
Last edited by KarishmaB on 18 Jan 2024, 01:54, edited 1 time in total.
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Re: The sum of 8 consecutive integers is 180. [#permalink]
adkikani wrote:
The sum of 8 consecutive integers is 180. What is the sum of first 4 of these integers?

a. 80
b. 82
c. 85
d. 86
e. 90

\(a + (a + 1) + (a + 2) + (a + 3) + (a + 4) + (a + 5) + (a + 6) + (a + 7) = 180\)

Or, \(8a + 28 = 180\)

Or, \(a = \frac{152}{8} = 19\)

Now, sum of \(a + (a + 1) + (a + 2) + (a + 3) = 4a + 6 = 4*19 + 6 => 82\), Answer must be (B)
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Re: The sum of 8 consecutive integers is 180. [#permalink]
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