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
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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.
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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.
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