In a certain quiz that consists of 10 questions, each question after

Hi,

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

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

\(10x + ( 4 + 8 + 12 + ... + 32+ 36) = 360\)

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, \(x + 8 = 26\)

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.

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

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.

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

Hi why is the last term 32 and not 36? should it be the 9th term which is x+36? thanks

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Yes. Edited the typo. Thanks!
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