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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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.

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 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

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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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

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

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.

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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