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The sum of the fourth and twelfth term of an arithmetic

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The sum of the fourth and twelfth term of an arithmetic [#permalink]

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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Sep 2012, 02:05, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Arithmetic progression [#permalink]

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New post 06 Aug 2009, 23:35
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OA C

Sum of the AP = avg of AP x no of terms
Avg of the AP can be found by addition of (first term + last term) /2 or (second +second last term)/2 or (third +third last term)/2.... try this out its a property of any AP....here as we need to find sum of first 15 terms...let us assume AP of 15 terms....

for this AP..going by above logic..... (1st +15th term) = (4th + 12 th term) = 20
hence Avg of AP = 10

Sum of the AP = 10 x 15 = 150 :-D
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Re: Arithmetic progression [#permalink]

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New post 06 Aug 2009, 23:54
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IMO C...

n th term of A.P. is given by a+(n-1)d
4 th term = a+3d
12 th term = a+11d

Given a+3d+a+11d=20 --> 2a+14d=20 --> a+7d = 10

Sum of n term of A.P = n/2[2a+(n-1)d]
subsitiuing n = 15 ...we get 15/2[ 2a + 14d] = 15 [a+7d] = 15*10 = 150...
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Re: Arithmetic progression [#permalink]

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Formula for Finding the nth term of an A.P. = Tn = a + (n-1) d

Using this we can calculate the the fourth term and the 12th term of this AP ..

T4 = a + 3d
T12 = a+11 d

We know that T4 + T12 = 20 a

Therefore ; a+3d + a + 11d = 20 ----> 2a + 14 d...

The question asks us to find the sum of the first 15 terms ..

Therefore using the formula for the Sum of n terms of an AP we get :

S15 = n/2 [ 2a + (15-1)d] [/b] ----> n/2 [ 2a + 14 d]

We have calculated the value for 2a + 14 d to be 20 , therefore we have S 15 = 15/2 (20) = 150 (C)
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Re: The sum of the fourth and twelfth term of an arithmetic [#permalink]

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For any Arithmetic Progression, the n'th term can be written as (a + (n-1)d), where 'a' = first term, 'd' = Common difference between two terms, and 'n' = number of terms.
e.g. 3rd term = a+2d, 1st term = a + 0d = a

The sum of the fourth and twelfth term of an arithmetic progression is 20.
4th Term + 12th Term = 20 ----------> (a+3d) + (a+11d) = 20 ----------> 2a + 14d = 20


What is the sum of the first 15 terms of the arithmetic progression?

Sum of n terms of a arithmetic progression is given by \(Sum = \frac{n(a+l)}{2}\), where 'n' = number of terms, 'a' = first term, and 'l' = last term

\(Sum of first 15 terms = \frac{15(1st term+15th term)}{2}\) ----------> \(\frac{15(a+(a+14d))}{2}\) ------------> \(\frac{15(2a+14d)}{2}\) ---------> \(\frac{(15)(20)}{2}\) -------------> \(150\)
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Re: The sum of the fourth and twelfth term of an arithmetic [#permalink]

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Re: The sum of the fourth and twelfth term of an arithmetic [#permalink]

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New post 11 Oct 2015, 20:36
Hello from the GMAT Club BumpBot!

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Re: The sum of the fourth and twelfth term of an arithmetic   [#permalink] 11 Oct 2015, 20:36
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