# Find the value of x such that 25 + 22 + 19 + 16 + …. + x = 112.

To Find: The value of x, i.e. the last term.

Given: The series and its sum.

The series can be written as x, (x + 3), …, 16, 19, 22, 25

Let there be n terms in the series

25 = x + (n - 1)3
3(n - 1) = 25 - x
x = 25 - 3(n - 1) = 28 - 3n

Let S be the sum of the series

n[28 - 3n + 25] = 224

n(53 - 3n) = 224

3n2 - 53n + 224 = 0

n = 7 as n cannot be a fraction.

Therefore, x = 28 - 3n = 28 - 3(7) = 28 - 21 = 7

The value of x is 7.

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