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