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


Answer :

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,

\(\Rightarrow \) 20.75 = 5 + (n - 1) (1.75)
\(\Rightarrow \) 20.75 = 5 + 1.75n - 1.75
\(\Rightarrow \)17.5 = 1.75n
\(\Rightarrow \) n = 10

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

NCERT solutions of related questions for Arithmetic Progressions

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