Asked by Estrick Abel
1)The sum of 6 terms of an arithmetic series is 45, the sum of 12 terms is -8. Find the first term and the common difference?
2)If in a geometric sequence the sum of the second and the third is 20 and the sum of the fourth and fifth is 320, find the common ratio and the first term. Assume that the common ratio is possitive.
2)If in a geometric sequence the sum of the second and the third is 20 and the sum of the fourth and fifth is 320, find the common ratio and the first term. Assume that the common ratio is possitive.
Answers
Answered by
Reiny
You have to know the formula for the sum of n terms of an AS
Sum(6) = 3(2a + 5d)
6a + 15d = 45 --- #1
sum(12) = 6(2a + 11d ) = -8
12a + 66d = -8
6a + 33d = -4 ---#2
subtract them
18d = -49
d = -49/18
back in #1
6a + 15(-49/18) = 45
6a = 45 + 735/18
6a = 515/18
a = 515/36
expected nicer numbers....
#2:
ar + ar^2 = 20 ---> ar(1 + r) = 20 ----#1
ar^3 + ar^4 = 320 ---> ar^3(1 + r) = 320 ---#2
divide #2 by #1
r^2 = 16
r = 4 , you said r is positive
in #1
4a(5) = 20
a = 1
a = 1 , r = 4
check:
terms would be
1 4 16 64 256
and 4+16 = 20
and 64 + 256 = 320
Sum(6) = 3(2a + 5d)
6a + 15d = 45 --- #1
sum(12) = 6(2a + 11d ) = -8
12a + 66d = -8
6a + 33d = -4 ---#2
subtract them
18d = -49
d = -49/18
back in #1
6a + 15(-49/18) = 45
6a = 45 + 735/18
6a = 515/18
a = 515/36
expected nicer numbers....
#2:
ar + ar^2 = 20 ---> ar(1 + r) = 20 ----#1
ar^3 + ar^4 = 320 ---> ar^3(1 + r) = 320 ---#2
divide #2 by #1
r^2 = 16
r = 4 , you said r is positive
in #1
4a(5) = 20
a = 1
a = 1 , r = 4
check:
terms would be
1 4 16 64 256
and 4+16 = 20
and 64 + 256 = 320
Answered by
Kenneth
the common different between p and q
Answered by
Anonymous
The third term of geometric sequence is 108 and the 6term is -32 find the common ratio ,first term and sum of first term
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