arithmetic progression

vavilaladivya wrote: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. 84

i am not able to solve this.. .... can someone please tell me how to go abt it ???

For such questions you need to remember that the sum of 4th and 12th term of an AP is the same as the sum of the 1st and 15th term. For example in the above case

t4 + t12 = 8 = t1 + t15
also known is that -average of this series of 15 terms will be equal to 1/2 ( sum of the first and the last term)


Average of t1 , t2, t3 and so on till t15 = 8/2 =4

sum of 15 terms = average of 15 terms x no of terms = 4 x 15 = 60

imo 60 is the answer

vavilaladivya wrote: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. 84

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