PreCalculus geometric series problems.?!?
Can someone help me out with ONE or BOTH of these problems please?! Thank you!
The sum of 10 + 15/2 + 5 + 5/2 + ... where the associated sequence has 30 terms.
How many terms of the sequence −5, −1, 3, ... must be added to give a sum of 400?
Can someone help me out with ONE or BOTH of these problems please?! Thank you!
The sum of 10 + 15/2 + 5 + 5/2 + ... where the associated sequence has 30 terms.
How many terms of the sequence −5, −1, 3, ... must be added to give a sum of 400?
-
both of these series are arithmetic.
10 + 15/2 + 5 + ...
1st term 10, common difference -2½,
so 30th term 10 – 2½(29) = 12½ – 2½(30) = 12½ – 75 = -62½
sum of 1st 30 terms is (10 – 62½)(30/2) = -52½(15) = -787½
-5 + -1 + 3 + ... =
common difference + 4
last term is -5 + 4(n – 1)
sum is 1st + last times n/2, so
[-10 + 4(n – 1)] (n/2) = 400
[ -10 + 4n – 4 ] n = 800
4n² – 14n = 800
2n² – 7n – 400 = 0
2n² – 32n + 25n – 400 = 0
2n(n – 16) + 25(n – 16) = 0
(2n + 25)(n – 16) = 0
n = 16 terms
10 + 15/2 + 5 + ...
1st term 10, common difference -2½,
so 30th term 10 – 2½(29) = 12½ – 2½(30) = 12½ – 75 = -62½
sum of 1st 30 terms is (10 – 62½)(30/2) = -52½(15) = -787½
-5 + -1 + 3 + ... =
common difference + 4
last term is -5 + 4(n – 1)
sum is 1st + last times n/2, so
[-10 + 4(n – 1)] (n/2) = 400
[ -10 + 4n – 4 ] n = 800
4n² – 14n = 800
2n² – 7n – 400 = 0
2n² – 32n + 25n – 400 = 0
2n(n – 16) + 25(n – 16) = 0
(2n + 25)(n – 16) = 0
n = 16 terms