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.
(a) the twentieth term of the arithmetic sequence
(b) the sum of the first twelve term of the G.P.
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.
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
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
2d = a (3 - 1)
2d = 2a
d = a
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