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
where a1 = first term = 7 and
d = common difference = 2, and n = number of term = 15
...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
where a1 = first term = 7 and an = a15 = 35.
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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,
But if you don't know the last term, then use the other one:
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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
We substitute Sn = 66 and solve for n
Multiply both sides by 2
Divide through by -3
Factor:
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