Q. 105.0( 2 Votes )

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

Answer :

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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