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

**Solution:**

aₙ = a + (n - 1)d is the nth term of AP.

Here, a is the first term, d is a common difference and n is the number of terms.

From the given data, Ramkali’s savings in the consecutive weeks are ₹ 5, ₹ (5 + 1.75), ₹ (5 + 2 × 1.75), ₹ (5 + 3 × 1.75) ... and so on

Hence, in n^{th} weeks the savings will be, ₹ [5 + (n - 1) × 1.75] = ₹ 20.75

According to the question, we have a = 5, d = 1.75, aₙ = 20.75

We know that the nth term of AP is aₙ = a + (n - 1)d

20.75 = 5 + (n - 1)1.75

15.75 = (n - 1)1.75

(n - 1) = 15.75/1.75

n - 1 = 1575/175

n - 1 = 9

n = 10

So, in the 10^{th} week Ramkali's saving will be ₹ 20.75.

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 5

**Video Solution:**

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

Class 10 Maths NCERT Solutions Chapter 2 Exercise 5.2 Question 20

**Summary:**

Ramkali saved ₹ 5 in the first week of a year and then increased her weekly saving by ₹ 1.75. If in the n^{th} week, her weekly savings become ₹ 20.75, the value of n is 10.

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