SOLUTION: the sum of the arithmetic serie whose first two terms are -12 and -8, respectively, and whose last term is 124.

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Question 1180187: the sum of the arithmetic serie whose first two terms are -12 and -8, respectively, and whose last term is 124.
Found 2 solutions by mananth, MathLover1:
Answer by mananth(16946) About Me  (Show Source):
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the sum of the arithmetic serie whose first two terms are -12 and -8, respectively, and whose last term is 124.
-12, -8, ...........124
a= -12, d=4, tn =124
tn = nth term = a+(n-1)d
124 = -12+(n-1)4
136 = 4n-4
140 = 4n
n= 35 There area 35 terms
Sum of terms Sn = (n/2) * (a+l) where a is the first term and l is the last term
Sn = (35/2) (-12+124)
= 35*112/2
35* 56 =1960
sum of the arithmetic series = 1960

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

the sum of the arithmetic serie whose first two terms are -12 and+-8, respectively, and whose last term is 124
a%5B1%5D=-12
a%5B2%5D=-8
last term is a%5Bn%5D=124
common difference is d=4
nth term formula is:
a%5Bn%5D=a%5B1%5D%2B4%28n-1%29....substitute given
124=-12%2B4%28n-1%29......solve for n
124%2B12=4%28n-1%29
136%2F4=%28n-1%29
34=n-1
34%2B1=n
n=35
there are 35 terms in this sequence, and they are
the sum is
sum%28-124%28n-+1%29%2C+1%2C35+%29+=+1960