Find the value of x such that

Solution:

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

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

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

Let S be the sum of the series

$S=\frac{n}{2}[x+25]=112$

$\Rightarrow \mathrm{n}[28-3 \mathrm{n}+25]=224$

$\Rightarrow \mathrm{n}(53-3 \mathrm{n})=224$

$\Rightarrow 3 \mathrm{n}^{2}-53 \mathrm{n}+224=0$

$\Rightarrow(n-7)\left(n-\frac{32}{3}\right)=0$

⇒ n = 7 as n cannot be a fraction.

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

The value of x is 7.