What is the first term of arithmetic sequence if the sum a1+a2+...a13=130? a4,a10,a7 are also consecutive terms of geometric sequence
Let the first term of the arithmetic sequence be a and the common difference be d.
(a + a + 12d)*(13/2) = 130
=> 2a + 12d = 20
=> a + 6d = 10
=> a = 10 - 6d
a4, a10 and a7 are consecutive terms of a geometric series:
=> (a + 9d) / (a + 3d) = (a + 6d) / (a + 9d)
=> (10 - 6d + 9d) / (10 - 6d + 3d) = (10 - 6d + 6d) / (10 - 6d + 9d)
=> (10 + 3d) / (10 - 3d) = 10 / (10 + 3d)
=> 100 + 9d^2 + 60d = 100 - 30d
=> 9d^2 + 90d = 0
=> d( d + 10) = 0
=> d = 0 and d = -10
So the first term of the series = 10 - 6*d can be 10 or 70
The required value of the first terms of the AP can be 10 or 70.
We'll write the sum of the first 13th terms of an arithmetical sequence:
S13 = (a1+a13)*13/2
130 = (a1+a13)*13/2
We'll divide by 13 both sides:
10 = (a1+a13)/2
a1 + a13 = 20
But a13 = a1 + 12d, where d is the common difference of the arithmetical sequence.
a1 + a1 + 12d = 20
2a1 + 12d = 20
a1 + 6d = 10 (1)
We also know that a4,a10,a7 are the consecutive terms of geometric sequence.
a10^2 = a4*a7
We'll write a4,a10,a7 with respect to a1 and the common difference d.
(a1 + 9d)^2 = (a1 + 3d)(a1 + 6d)
We'll expand the square and we'll remove the brackets:
a1^2 + 18a1*d + 81d^2 = a1^2 + 9a1*d + 18d^2
We'll eliminate a1^2, we'll move all terms to one side and we'll combine like terms:
9a1*d + 63d^@ = 0
We'll divide by 9:
a1*d + 7d^2 = 0
We'll factorize by d:
d(a1 + 7d) = 0
a1 = -7d (2)
We'll substitute (2) in (1):
-7d + 6d = 10
-d = 10
d = -10
a1 = 10 - 6d
a1 = 10 + 60
a1 = 70
The first term of the arithmetical progression is a1 = 70, for d = -10 and a1 = 10 for d = 0.