# 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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### 2 Answers

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.**