Arithmetic sequence question

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For an arithmetic sequence, let a2 = 15 and a10 = 10
Find: d, a1

In arithmetic series, general term, a_n is given by

a_n = a_1 + (n - 1)d

a_2 = a_1 + d

=> a_1 + d = 15 -------------eqn(1)

a_10 = a_1 + 9d

=> a_1 + 9d = 10 ----------eqn(2)

subtract eqn(1) from eqn(2)

=> 8d = -5

d = -5/8

substitute in eqn(1)

a_1 - 5/8 = 15

a_1 = 15 + 5/8

= 125/8

If the first term of an arithmetic sequence is a, and the common difference is d, then the arithmetic sequence would look like this:

a, a+d, a+2d, a+3d, a+4d, ...

Thus, the nth term of an arithmetic sequence is: a + (n - 1)d

It is given that a(2) = 15 and a(10) = 10. For a(2), n = 2 and for a(10), n = 10.
So,
a(2) = a + (2 - 1)d = 15
=> a + d = 15 ........(1)

a(10) = a + (10 - 1)d = 10
=> a + 9d = 10 .....(2)

Subtracting left and right sides of (1) from (2), we have

(a + d) - (a + 9d) = 15 - 10
=> -8d = 5
=> d = -5/8

From (1), a + (-5/8) = 15 => a = 15 + (5/8) = 125/8.

First term a(1) = 125/8
Common difference d = -5/8