Question

A man saved 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. His total saving from the start of service will be Rs. $11040$ after.

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

Answer 1

The correct option is B $19$ months

$\frac{\stackrel{1st}{Month}}{200}\frac{\stackrel{2nd}{Month}}{200}\frac{\stackrel{3rd}{Month}}{200}\frac{\stackrel{4th}{Month}}{240}\frac{\stackrel{5th}{Month}}{280}$

So let $n$ be the number of monthe.

Final savings $=11040$

So $\underset{}{200+200}+200+240+280+....=11040$.

$\Rightarrow 200+240+280+...+200+(n-1)40=11040-400=10640$

Here $a=200,d=40$.

We know $a+(a+d)+....+af(n-1)d=\frac{n}{2}(2a+(n-1\left)d\right)$

$\Rightarrow \frac{n}{2}(2\times 200+(n-1)\times 40)=10640$

$\Rightarrow n(n+q)=532$

$\Rightarrow n^2+qn-532=0$.

$\Rightarrow (n+28)(n-19)=0$.

$\Rightarrow n=19$ (Since $n=-28$ not possible)