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arithmetic progression

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by MBA.Aspirant » Wed Nov 23, 2011 9:19 am
The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression?

A. 300
B. 120
C. 150
D. 170
E. 270
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by vishal.pathak » Wed Nov 23, 2011 9:57 am
MBA.Aspirant wrote:The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression?

A. 300
B. 120
C. 150
D. 170
E. 270
Nth term of AP = a + (n-1)d
Sum of an AP = n/2(2a + (n-1)d)

Sum of 4th and 12th term = a +(4-1)d + a + 11d = 2a +14d = 20 (given)

Sum of 1st 15 terms = 15/2( 2a + (15 - 1)d ) = 15/2 (2a + 14d) = 15/2 *20 = 150 IMO C
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by Anurag@Gurome » Wed Nov 23, 2011 7:32 pm
MBA.Aspirant wrote:The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression?

A. 300
B. 120
C. 150
D. 170
E. 270
nth term of an AP, a(n) = a + (n - 1)d, where a = 1st term, n = no. of terms, and d = common difference.
Now, 4th term, a(4) = a + (4 - 1)d = a + 3d
and 12th term, a(12) = a + (12 - 1)d = a + 11d
Given: a(4) + a(12) = 20
a + 3d + a + 11d = 20
2a + 14d = 20...Equation (1)

Now sum of n terms of an AP, S(n) = (n/2) * [2a + (n - 1)d]
So, S(15) = (15/2) * [2a + (15 - 1)d] = (15/2) * [2a + 14d] = (15/2) * [20]...[putting the value of 2a + 14d from equation 1]
Simplifying, we get S(15) = = (15/2) * [20] = 150

The correct answer is C.
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by GMATGuruNY » Thu Nov 24, 2011 3:54 am
MBA.Aspirant wrote:The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression?

A. 300
B. 120
C. 150
D. 170
E. 270
The solution here is very quick and involves very little math if we understand a few basic facts about arithmetic progressions.

An arithmetic progression is a set of EVENLY SPACED NUMBERS.
Given a set of evenly spaced numbers, SUM = NUMBER OF TERMS * MEDIAN.
The median of the 15 terms above is the 8th term, which is the AVERAGE of the 4th and 12th terms.
Since the sum of the 4th and 12th terms is 20, their average (and the MEDIAN of the entire set) = 20/2 = 10.
SUM = NUMBER OF TERMS * MEDIAN = 15*10 = 150.

The correct answer is C.
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by chieftang » Tue Nov 29, 2011 2:18 pm
GMATGuruNY wrote:
MBA.Aspirant wrote:The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression?

A. 300
B. 120
C. 150
D. 170
E. 270
The solution here is very quick and involves very little math if we understand a few basic facts about arithmetic progressions.

An arithmetic progression is a set of EVENLY SPACED NUMBERS.
Given a set of evenly spaced numbers, SUM = NUMBER OF TERMS * MEDIAN.
The median of the 15 terms above is the 8th term, which is the AVERAGE of the 4th and 12th terms.
Since the sum of the 4th and 12th terms is 20, their average (and the MEDIAN of the entire set) = 20/2 = 10.
SUM = NUMBER OF TERMS * MEDIAN = 15*10 = 150.

The correct answer is C.
Very elegant solution. I'm no expert, but I would strongly suspect that no GMAT problem actually requires memorization of arithmetic progression formulas to solve. Do you think that is a fair assumption?

Thanks!
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