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by gmatjedi » Fri May 28, 2010 3:17 am
a(n)=a(1)+(n-1)d

a(4)=a(1)+3d
a(12)=a(1)+11d

a(4)+a(12)=20

20=2a(1)+14d

10=a(1)+7d
therefore a(1)=3 and d=1

start sequence at 3...17

sum of first 15 terms= (3+17)/2 * 15= 150
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by liferocks » Fri May 28, 2010 4:23 am
T(4)=a+3d
T(12)=a+11d
T(4)+T(12)=2a+14d=20

S(15)=15/2*(2a+14d)=15/2 * 20=150

Ans option C
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by selango » Fri May 28, 2010 8:15 am
AP formula =n/2[2a+(n-1)d]


nth term = a+(n-1)d

4th term =a+3d

12th term=a+11d


-->a+3d+a+11d=20
2a+14d=20

Substituting in AP formula with n=15

15/2[2a+14d]


15/2*20

=150
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by odod » Fri May 28, 2010 12:55 pm
Hello.

I need some help understanding this solution.

what is a(n)=a(1)+(n-1)d ?. Is that formula to know for the gmat. If so, can someone please explain it to me?

I would appreciate any help.

Thanks
ODOD
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by kevincanspain » Fri May 28, 2010 1:34 pm
gmatmachoman 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 the 4th and 12th term is 20, so is the sum of the 1st and 15th. Thus we have 15 terms that have a mean of 20/2 = 10. Sum= 150
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by HPengineer » Fri May 28, 2010 1:42 pm
Here a link to the formula used from wikipedia... Might shed some light

https://en.wikipedia.org/wiki/Arithmetic_progression
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