Question

A man saves $Rs.200$ in each of the first three months of his service. In each of the subsequent months his saving increases by $Rs.40$ more than the saving of immediately previous month. After how many months his total savings from the start of service will be $Rs.11040$.

• A. $18$ months
• B. $19$ months
• C. $20$ months
• D. $21$ months

Answer 1

Answer: (D)

$21$ months

Savings are:

$200,200,200,240,280,.......\text{up to n terms}$

But $200,240,280,......(n-2)$ terms forms an A.P

Since, sum of n terms of an AP $=S_n=\frac{n}{2}[2a+(n-1\left)d\right]$, where first term is $a$ and common difference $d$

$\therefore$ According to question

$400+\frac{n-2}{2}\left[2\right(200)+(n-2-1\left)40\right]=11040$ ...... Sum $=11040$

$\Rightarrow 400+(n-2)[200+(n-3\left)20\right]=11040$

$\Rightarrow (n-2)[200+20n-60]=10640$

$\Rightarrow (n-2)(20n+140)=10640$

$\Rightarrow 20(n+7)(n-2)=10640$

$\Rightarrow n^2+5n-14=532$

$\Rightarrow n^2+5n-546=0$

$\Rightarrow (n+26)(n-21)=0$

$\Rightarrow n=21$

So, total time $=21$ months.