Hi,

If we look at the increase, it tells us that it is a arithmetic expression..

let the first number be x..
so \(SUM \ \ of \ \1st \ \ 10= x+4+x+x+8+....x+4*9\)..
so \(SUM = 10x+4(1+2+..+9)=360\)

10x+180=360..
10x=180 or x=18..

so 3rd number= 18+4*2=26
C
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Equation is like this:

\(10x + ( 4 + 8 + 12 + ... + 32) = 360\)

to find sum of the evenly spaced sequence, use average * number of the members in the sequence

\(10x + (20*9) = 360\)

\(10x + 180 = 360\)
\(x = 18\)

So, \(x + 8 = 26\)

chetan went the orthodox way.
But Aves chose the short cut and ended up creating confusion in middle of the solution. i understand that very well.
What you did further was like you forgot what you said above.

Thanks for the detailed explanation, helped a lot. But the red 32 has to be a 36, right?

Good shortcut - will use this in the future.

I did it the long way but still got the question right in about 90 seconds

x
x+4
x+8
x+12
x+16
x+20
x+24
x+28
x+32
x+36

10x+180=360
10x=180
x=18

3rd question = x+8 = 18+8 = 26

Answer C

I found a shortcut for the same. Since we know that there are 10 questions on the quiz and all questions are evenly placed by 4 points, DO NOT choose the points in the first question as x.

INSTEAD, choose it as x-16.
So, the numbers will be:
1) x-16
2) x-12
3) x-8
4) x-4
5) x
6) x+4
7) x+8
8) x+12
9) x+16
10) x+20

Adding all the above 10 will lead to a simple sum of 10x + 20 (-16 & 16 will lead to a sum of 0, same for -12& +12 so on & so forth)

SO, 10x + 20 = 360
10x= 340
x= 34
3rd question = x-8= 34-8 = 26.

Hope it helps.

I did it this way also.

The quick way to do this math is
x + 4(sum of 1 through 9) =
x + 4(45) =
x + 180

(Sum of 1 through 9) is found by:
Average of first and last term * number of terms =
1 + 9 = 10/2 = 5
1 through 9 = 9 terms
9*5 = 45

This is an Arithmetic Progression (A.P.) with terms => a, a+4, a+8, ...., a + (n-1) *4

a => First Term of Series
n => Number of the term
l => Last Term

So, l = a + (10-1) * 4 = a + 9*4 = a + 36

Sum of an A.P. Series = \(\frac{n*(a+l)}{2}\) => Standard Formula

Therefore,

\(360 = \frac{n*(a+l)}{2} => 360 = \frac{10*(a+a+36)}{2}\)

=> 360 = 5 * (2a + 36) => 72 = 2a + 36

=> 2a = 36 => a = 18

Therefore, 3rd Term = a + 8 = 26
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I got this in a similar approach but we can calculate and get answer in 50 Secs.

Since the seq is in AP evenly spaced the sum= Avg* N

Seq is x,x+4, x+2*4,x+3*4, x+4*4,x+5*4,x+6*4,x+7*4,x+8*4,x+9*4 ( don't need to write the entire seq just get the idea how seq is going to form )

and so on the average of evenly spaced set of 10 numbers is average of 5th and 6th value.

Since each term is increment of 4 the mid value between 5 and 6 will be increment of 2 from 5th value
so fifth value is x+4*4= x+16
so average is x+18 so
(x+18)*10=360
x+18=36
or x=18
and third value is x+2*4=18+8= 26

The trick in above question is identifying the evenly spaced seq, then its average or median value and finding x.
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since we have 10 items on this list, starting with n and ending with n+36, I used empowergmat's bunching method: (note that this is useful for lists that have consecutive forms, in this case with +4 increments)

method: Add the first and Last number, then multiply it by half of the number of the items, in this case 5 (given our 10 item list)

so, (n + n + 36) * 5 should equal to 360 ---> 2n + 36 = 72 --> 2n = 36 ---> n = 18
since we are looking for 3rd item, that's n + 8, which is 18 + 8 = 26

EMPOWERgmatRichC please let me know if I used the method correctly... not sure if I had ever used bunching in a sequence question like this, but it def appeared to have worked.

This is an Arithmetic progression question.

n/2 * [2a + (n - 1) *d] = 360
10/2 * [2a + (10 - 1)* 4)] = 360
a = 18

So, first question is 18 marks.

So, third question = 26

The first question is worth x, the 2nd worth x + 4, the 3rd worth x + 2(4) = x + 8, ..., the 10th is worth x + 9(4) = x + 36.

Since we have a set of evenly-spaced numbers, the average is (x + x + 36)/2 = (2x + 36)/2 = x + 18.

Since sum = average * quantity we have:

360 = (x + 18) * 10

36 = x + 18

18 = x

The first question is worth x = 18 points. So the third question is worth x + 8 = 18 + 8 = 26 points.

Answer: C
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10x + 4 (1 + 2 + 3 + ....9) = 360

=) 10x + 4 * [n(n+1)]/2 = 360

=) 10x + 4 * [(9 * 10)] / 2 = 360

=) x = 18

third question = (18 +8) =26 = C the answer

thakns

There is no need for any formulas here. Sum of 10 consecutive integers = 360. Average = 36. Average of a set of consecutive integers = Median of the set. Median is the number between 5 and 6, and its equals 36.

Since the difference between terms is 4, therefore, 5th term is 34 and 6th term is 38.
5th term = 34
4th term = 30
3th term = 26

If we assume first question is worth "x" points, the proceeding questions would be worth x+4, x+8, x+12,.... x+36 points.

So, the sum of 10 questions would be 10x plus the sum of 4 - 36 inclusive, with multiples of 4 (4, 8, 12, 16, 20, 24, 28, 32, 36).

The sum of 4-36 can be calculated with (4+36)/4*(16+20)/2 = 180. (The first and last numbers in the list divided by 4, times the average of the numbers (16+20)/2=18)

Therefore, the total sum would be 10x + 180 = 360, x= 18.

Finally the third question was worth x+8 = 18+6 = 26 points.

Let the first question is worth ‘a’ points.
Then, Second question is worth a + 4 points, third of worth (a + 4 + 4) = a + 8 points.
So, the sum of the series becomes a + a + 4 + a + 8 + a + 12 ……… upto 10 terms
= 10a + 4 + 8 + 12 + ……. upto 9 terms
Sum = 10a + Sum of A.P with first term 4 and common difference 4.
Sum = 10a + 180 = 360
a = 18
Third question is worth a + 8 points = 18 + 8 = 26 points.

Hi,
here are my two cents for this question

First if we recognize that this is a AP with common difference 4 then we can use the concept of corresponding term

Using the concept of corresponding terms , we can say that average f corresponding terms * Number of terms =sum of all terms

Here is what i mean

So

Number of terms is 10
\(a_1\), \(a_2\), \(a_3\), \(a_4\), \(a_5\), \(a_6\), \(a_7\), \(a_8\), \(a_9\), \(a_{10}\)

Then corresponding terms will be
\(a_1\) & \(a_{10}\) , \(a_2\) & \(a_9\) , \(a_3\) & \(a_8\) , \(a_4\) & \(a_7\) , \(a_5\) & \(a_6\)


Since we are asked to find the value of \(a_3\) i will take the corresponding pair of \(a_3\) & \(a_8\)

Also since its an AP series we have \(a_8\)= \(a_3\)+20

So
360= 10 * \(\frac {a_3 + a_8}{2}\)

so substituting relevant information we have

360= 10 * \(\frac {2(a_3) + 20}{2}\)

72= 2\(a_3\) + 20


2\(a_3\) = 52

\(a_3\)= 26

Probus
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\(10a + (\frac{(4+36)}{2 }* \frac{(36-4)}{4} +1) = 360\)

\(a = 18\)

\(a_3 = 26\)

Bunuel this question needs the Gmatprep tag, thanks!

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