If the sum of the 4th term and the 12th term of an arithmetic progression is 8, what is the sum of the first 15 terms of the progression?

A. 60
B. 120
C. 160
D. 240
E. 840

Kudos for a correct solution.
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IMO: A

nth term of an AP = a+(n-1)d
Let 1st term of AP = a+d
12th term of AP = a+11d
Sum of 1st & 12th = 8
a+d+a+11d =8
2a+14d=8
a+7d = 4 ---(i)

Sum of 1st 15 terms =
(a+d)+(a+2d)+(a+3d)+.....(a+14d)
=15a+(d+2d+3d+4d+..14d)
= 15a+d(1+2+3+..14)
Sum of 1st n natural numbers = \(\frac{n(n+1)}{2}\)
= 15a +d (\(\frac{14*15}{2}\))
= 15a +d(7*15)
=15(a+7d) --(sub eq---(i))
=15*4=60
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And yes, I like kudos ¯\_(ツ)_/¯

First term=a, common difference=d
4th term=a+3d
12th term=a+11d
Sum=2a+14d=8

Now , sum upto first 15 terms= 15/2(2a+14d)=15/2(8)=60
Answer A

Ans: A

solution: given that, lets say first term is a and common difference is d
then, a+3d+a+11d=8 = a+7d =4 ===this is the 8th term, which will be the median of the set of 15 terms in AP
so the sum of all the terms in AP= number of terms*median = 15*4 = 60
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t4+t12=8
in AP 1,2,3,4,5.., where a and d=1, t4+t12=2*8
if we assume a and d=1/2,
then ∑15=15/2*(1+14*1/2)=60

An arithmetic progression (sequence) is characterized by having each term separated from the next term by a common difference. We can let d = the common difference (i.e., the difference between each pair of consecutive terms) and let the first term = a.

Thus, the first term is a, second term is a + d, and third term is a +2d, so the 4th term is a + 3d and the 12th term is a + 11d. Thus:

(a + 3d) + (a + 11d) = 8

2a + 14d = 8

a + 7d = 4

We are asked to find the sum of the first 15 terms. Since this is an arithmetic progression, we can use the formula sum = quantity x average. The ‘quantity’ is 15 since there are 15 terms. The ‘average’ of a finite arithmetic progression is also the median, which in this case is the 8th term. The 8th term in terms of a and d is a + 7d, and we have found that to be 4. Thus:

Sum = 15 x 4 = 60

Answer: A
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