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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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New post 27 Sep 2012, 21:12
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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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New post 16 Mar 2017, 23:49
I'm sorry if this is not germane to the post but in what particular math class or type of math do you learn these formulas such as the sum of terms in a series? I have done elementary-calculus and am now in a college level statistics course. I have not seen any of these series formulas.

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

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New post 17 Mar 2017, 00:55
joyseychow 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

let a be the first term n be the number of terms and d be the difference

a+3d +a +11d = 20 or
2a +14d = 20

sum of 15 terms = n/2{2a + (n-1)d}
= 15/2(2a +14d)
substituting we get

15*20/2 =150.

option C

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

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New post 19 Apr 2017, 07:41
Let fourth term be t4 and 12th term be t12.

Now t12 can be written as : t12 = t4 + 8d; ---- (Equation I)

Given : t4 + t12 = 20 ---- (Equation II)

Substituting value of t12 in Equation II, we get :

t4 + t4 + 8d = 20

=> t4 + 4d = 10

Which is nothing but t8. (Since t4 + 4d = t8)

We now the value of t8;

Also, 15 terms in AP can also be written as :

a - 7d, a - 6d, a - 5d, a - 4d, a - 3d, a - 2d, a - d, a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, a + 6d, a + 7d ; ---- (Equation III)

If you notice, a is nothing but the 8th term in above series.

Taking sum of all the terms in equation III, we get,

S15 = 15a = 15 * t8 = 15 * 10 = 150

which is the required answer. Option C

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

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New post 19 Apr 2017, 12:57
joyseychow 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


if the sum of fourth and twelfth terms were 16,
then first term and difference would both be 1
20/16=5/4
let 5/4*1=first term and difference
(5/4+15*5/4)/2=10
15*10=150
C

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

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New post 25 Aug 2017, 03:59
Bunuel,

Which approach is more preferable while solving such questions?
1.Sum of Ap = Average of Ap * No. of terms.

or

2. Converting them like
4 th term = a+3d
12 th term = a+11d
andthen adding them up?

I am asking this because I am not sure about the first approach.

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Re: The sum of the fourth and twelfth term of an arithmetic   [#permalink] 25 Aug 2017, 03:59
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