Find the difference of the arithmetic progression and its first term x10 = 1 S16 = 4.

In an arithmetic progression, any n-th term can be calculated as

x (n) = x (1) + d * (n – 1).

By the condition x (10) = a (1) + d * (10 – 1) or

a (1) + 9d = 1 – the first equation with two unknowns.

Sum of the first n terms of an arithmetic progression

S (n) = (2 a (1) + d * (n – 1)) * n / 2.

S (16) = 4 = (2 a (1) + d * (16 – 1)) * 16/2.

Second equation: 16 a (1) + 120d = 4.

By substituting a (1) from the first into the second equation and after transformations, we obtain

4 * (1 – 9d) + 30 d = 1.

Whence d = 0.5