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

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.