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Fast approach again needed

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Fast approach again needed

by bacali » Mon Dec 01, 2008 10:17 pm
I added 4+8+12+16+20....9x.... Took a while. Can someone set it up for a fast solve?

In a certain quiz that consists of 10 questions, each question after the first is worth 4 points more than the preceding question. If the 10 questions on the
quiz are worth a total of 360 points, how many points is the third question worth?

A. 18
B. 24
C. 26
D. 32
E. 44


OA: C
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Re: Fast approach again needed

by sudhir3127 » Mon Dec 01, 2008 10:23 pm
bacali wrote:I added 4+8+12+16+20....9x.... Took a while. Can someone set it up for a fast solve?

In a certain quiz that consists of 10 questions, each question after the first is worth 4 points more than the preceding question. If the 10 questions on the
quiz are worth a total of 360 points, how many points is the third question worth?

A. 18
B. 24
C. 26
D. 32
E. 44


OA: C
this is Arithemetic progression

Let first term = X
D = 4

360= 10/2 ( 2x+ 9*4)..............( Sn= n/2 ( 2a+ ( n-1)*d )

Solve for X , X= 18

thus the 3 term

18 + 4*2 = 26

Thus C
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by cramya » Mon Dec 01, 2008 10:24 pm
Treat it as a arithmetic sequence wiht a common difference of 4

Find a1 the first element of the sequence and add 8 to find the third item

Sum of n terms in an arithmetic sequence = n/2(2a1+(n-1)d)

where n->number of terms(here n =10)
a1->first term in the sequence (we need to find)
d-> common difference(here d= 4)

10/2 (2a1+(10-1)4) = 360
2a1+36 = 72
2a1 = 36
a1=18


a3 = 18+8 = 26

Done C)
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by vivek.kapoor83 » Tue Dec 02, 2008 3:30 am
same as cramya
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