T4 + T12 = 20

Sum of first 15 terms = 15/2 * (T1 + T15)
= 15/2 * (T4 - 3d + T12 + 3d)
= 15/2 *(T4 + t12)
Ans IMO = 150 i.e. C

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

benefit of a PS question.we can substitute anything as long as the numbers satisfy the condition

let's say all the numbers are 10.
so sum of first 15 numbers=150:)

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

Concept:
In general T1+Tn=T2+T(n-1)=T3+T(n-2) and so on. Please note that you can form this series for any of the middle number blindly. Here is how. In T1+Tn If you add 1+n it will be equal to 2+n-1 =n+1

In an arithmetic series T1+T15 = T2+14=T3+T13=T4+T12 and so on. Here 1+15=2+14=4+14.

We know that sum of 4th and 12th term is 20 hence average will be 20/2 =10. Therefore total will 10*15=150

Hence C

It's an 15 secs problem if you know the concept. Since I knew the concept prehand all that I need to do was divide 20/2 and multiply by 15.

Thanks,
Quant-Master