Question

Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the $n^{th}$ week , her weekly savings become Rs 20.75, find n.

Answer 1

Ramkali saved Rs. 5 in the first week of year.

It means first term = a = 5

Ramkali increased her weekly savings by Rs 1.75

Therefore, common difference = d = Rs 1.75

Money saved by Ramkali in the second week = a+ d = 5 + 1.75 = Rs 6.75

Money saved by Ramkali in the third week = 6.75 + 1.75 = Rs 8.5

Therefore, it is an AP of the form: 5, 6.75, 8.5... 20.75

We want to know in which year her weekly savings become 20.75.

Using formula $a_n=a+(n-1)d$, to find $n^{th}$ term of arithmetic progression, we can say that

$20.75=5+(n-1)\left(1.75\right)$

$\Rightarrow 20.75=5+1.75n-1.75$

$\Rightarrow 17.5=1.75n$

$\Rightarrow n=\frac{17.5}{1.75}=10$

It means in the ${10}^{th}$ week her savings become Rs 20.75.