Question 1191984: (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, greenestamps:
Answer by Boreal(15235)   (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
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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
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This is n(n+1) where n=25
The answer is 650.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The response from the other tutor showed formal algebraic solutions to these problems.
While you should know the formal algebra, plugging numbers into formulas doesn't do much to give you an understanding of the mathematics. Solving the problems in a less formal manner using logical reasoning can be helpful in building that understanding.
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n-th term of an arithmetic sequence with first term a and common difference d
Formally: t(n)=a+(n-1)d
Informally: The n-th term is the first term, plus the common difference (n-1) times.
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Sum of the first n terms of an arithmetic sequence with first term a and common difference d
Formally (if the n-th term is not known): S(n)=(n/2)(2a+(n-1)d)) <-- (ugly...!)
Formally (if the n-th term is known): S(n)=(n/2)(first+last)
Informally: S(n) = (number of terms)*(average of first and last terms)
Note the last formula is based on the trivial fact that the sum of any group of numbers is the product of the number of numbers and the average of the numbers; and in an arithmetic sequence the average of all the numbers is the average of the first and last.
We can rewrite the second formal formula above to say the calculations are exactly those in the informal method: S(n)=n((first+last)/2)
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With that introduction, now on to your questions -- solving them informally.
(a) (part 1) 15th term in an arithmetic sequence with first term 7 and common difference 2:
7, plus the common difference 14 times
7+14(2)=7+28=35
ANSWER: 15th term is 35
(a) (part 2) Sum of first 15 terms:
# of terms: 15
average of first and last: (7+35)/2=21
sum: 15*21=315
ANSWER: sum of first 15 terms is 315
(b) # of terms in the sequence -9, -6, -3, ... to make a sum of 66
We know the sum is 66; and we know that sum is the product of the number of terms and the average of the first and last terms. We also know that each term is a multiple of 3; that means the sum is a multiple of 3, and the average is also probably a multiple of 3.
So we can find the answer by looking at pairs of integers with one of them a multiple of 3 and the product 66:
22 terms with an average of 3? No; doesn't work with the given sequence
11 terms with an average of 6? Yes; it works. The 11th term would be -9+10(3)=21; and the average of first and last terms is (-9+21)/2=6; product is 11*6=66
ANSWER: 11 terms make the sum 66
(c) sum of first 25 even integers
# of terms: 25
average of 1st and 25th terms: (2+50)/2=26
sum: 25*26=650
ANSWER: 650
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