The first, third and ninth terms of a arithmetic sequence (A.P) are equal to the first three terms of a geometric sequence (G.P). if the seventh term of the linear sequence is 14.
Calculate:
(a) the twentieth term of the arithmetic sequence
(b) the sum of the first twelve term of the G.P.
+23245 Let the first term of the arithmetic sequence = a
and the common difference of the arithmetic sequence = d.
Let the first term of the geometric sequence = a
and the common ratio of the geometric sequence = r.
By some quick guessing, I came up with these values: a = 2 d = 2 r = 3.
From these, you can calculate the vaues you desire.
+128475 Let the first term of the arithmetic series = a
Let the third term = a + 2d
Let the ninth term = a + 8d
And we know that a + 6d = 14
The second term of the geometric series = ar
The thrid term of the geometric series = ar^2
So
a + 2d = ar ⇒
2d = ar - a ⇒
2d = a ( r - 1)
a + 8d = ar^2 ⇒
8d = ar^2 - a ⇒
8d = a (r^2 - 1) ⇒
8d = a (r- 1) (r + 1) ⇒
8d = (2d)(r + 1) ⇒
8 = 2 (r + 1) ⇒
4 = r + 1
So r = 3
And
2d = a (3 - 1)
2d = 2a
d = a
So
a + 6d = 14
a + 6a = 14
7a = 14
a = 2 so d also = 2
(a) The twelveth term of the arithmetic series = a + 11d = 2 + 11(2) = 24
(b) The sum of the first 12 terms of the geometric series is
S = a [ 1 - r^12 ] [ 1 - r ] = 2 [ 1 - 3^12] / [ 1 - 3 ] = [ 3^12 - 1 ] = 531440
