SOLUTION: a) The first term of an arithmetic progression is 7 and the common difference is 2. Find the 15th term and the sum of the first 15th term. (b) how many terms of the sequence -

Algebra ->  Finance -> SOLUTION: a) The first term of an arithmetic progression is 7 and the common difference is 2. Find the 15th term and the sum of the first 15th term. (b) how many terms of the sequence -      Log On


   

Question 1191995: a) The first term of an arithmetic progression is 7 and
the common difference is 2. Find the 15th term and the sum
of the first 15th term.
(b) how many terms of the sequence -9,-6,-3 must be taken
that the sum maybe 66.
(c) find the sum of the first 25 even integers.
Found 2 solutions by Boreal, Edwin McCravy:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
sum=(n/2)(2a+(n-1)d)
nth term =a1+(n-1)d
a. n15=7+(15-1)2=35
Sum is (15/2)(14+14*2)=315
7+9+11+13+15+17+19+21+23+25+27+29+31+33+35=315
-
a1=-9, d=3, sum=66
66=(n/2)(-18+(n-1)3)=(n/2)(-18+3n-3)=(n/2)(3n-21)
multiply by 2 to clear fractions
132=3n^2-21n
n^2-7n-44=0 dividing by 3 and rearranging.
(n-11)(n+4)=0
n=11 terms only positive root.
-9,-6,-3,0,3,6,9,12,15,18,21
-
This is n(n+1) where n=25
The answer is 650.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
a) The first term of an arithmetic progression is 7 and
the common difference is 2. Find the 15th term...
 
The "hard" way is to write them out starting with 7 and 
adding 2 each time
until you have 15 terms:

7,9,11,13,15,17,19,21,23,25,27,29,31,33,35.

The "easy" way is to substitute in the formula a%5Bn%5D=a%5B1%5D%2B%28n-1%29d
where a1 = first term = 7 and 
d = common difference = 2, and n = number of term = 15

a%5Bn%5D=a%5B1%5D%2B%28n-1%29d
a%5B15%5D=7%2B%2815-1%292
a%5B15%5D=7%2B%2814%292
a%5B15%5D=7%2B28
a%5B15%5D=35
...and the sum of the first 15th term.

The "hard" way is to add them up:

7+9+11+13+15+17+19+21+23+25+27+29+31+33+35 - 315.

The "easy" way is to substitute in the formula S%5Bn%5D=expr%28n%2F2%29%28a%5B1%5D%2Ba%5Bn%5D%29
where a1 = first term = 7 and an = a15 = 35.

S%5Bn%5D=expr%28n%2F2%29%28a%5B1%5D%2Ba%5Bn%5D%29
S%5B15%5D=expr%2815%2F2%29%287%2B35%29
S%5B15%5D=expr%2815%2F2%29%2842%29
S%5B15%5D=315%29

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There are two formulas for the sum of an arithmetic progression.

If you know the last term, it's easier to use the one we just used,

S%5Bn%5D=expr%28n%2F2%29%28a%5B1%5D%2Ba%5Bn%5D%29

But if you don't know the last term, then use the other one:

S%5Bn%5D=expr%28n%2F2%29%282a%5B1%5D%2B%28n-1%29d%29

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(b) how many terms of the sequence -9,-6,-3 must be taken that
the sum maybe 66.

We don't know the last term, so we use the second formula, with
a1=-9 and common difference = 
(2nd term) - (1st term) =  (-6) - (-9) = -6+9 = 3

S%5Bn%5D=expr%28n%2F2%29%282a%5B1%5D%2B%28n-1%29d%29

We substitute Sn = 66 and solve for n

66=expr%28n%2F2%29%282%28-9%29%2B%28n-1%29%283%29%29
66=expr%28n%2F2%29%28-18%2B3n-3%29
66=expr%28n%2F2%29%28-21%2B3n%29
Multiply both sides by 2
132=n%28-21%2B3n%29
132=-21n%2B3n%5E2%29
-3n%5E2%2B21n%2B132=0
Divide through by -3
n%5E2-7n-44=0
Factor:
%28n%2B4%29%28n-11%29=0
n+4=0;  n-11=0
  n=-4;    n=11

n can't be negative, so n = 11. It takes 11 terms to have sum 66.

Edwin