Find the sum of $n$ term of the given AP $0.9, 0.91, 0.92, 0.93...............$?

Question

First term $a= 0.9$, Common difference $d=0.01$ $$S_{n} = \frac n 2[2a+(n-1)d]$$ $$=\frac n 2[2(0.9)+(n-1)(0.01)]$$ $$=n(n+179)/200$$

Now my question is how we get $=n(n+179)/200$. Can explain how we get this answer. Thank you.

Answer 1

Basically we have $$ \frac{n}{2}[2(0.9) + (n-1)(0.01)]$$ $$= \frac{n}{2}[1.8 + 0.01n-0.01]$$ $$= \frac{n}{2}[1.79 + 0.01n]$$ $$=\frac{n}{2}\times \frac{1}{100}[179+n]$$ $$=\frac{n(n+179)}{200}$$ Hope it helps.

Answer 2

Hint: by the sum of an arithmetic progression $a_1=0.9, a_n=a_1+ (n-1)\cdot 0.01$ so: $$S_n=n\,\frac{a_1+a_n}{2}=n \frac{0.9+0.9+(n-1)\cdot 0.01}{2} = n \frac{180 + (n-1)}{200}=\cdots$$