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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
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Bhushan S. If you like my post....Consider it for Kudos

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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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.

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

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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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