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Q20:
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the $n$th week, her weekly savings become ₹ 20.75, find $n$.

Solution :

Given:

Ramkali's savings in the first week ($a_1$) = ₹ $5$.

The weekly increase in savings ($d$) = ₹ $1.75$.

The savings in the $n$th week ($a_n$) = ₹ $20.75$.

To Find:

The value of $n$ (the number of weeks).

Step 1: Identifying the Arithmetic Progression (AP)

Since the savings increase by a constant amount each week, the sequence of savings forms an Arithmetic Progression.

Let the first term be $a = 5$.

Let the common difference be $d = 1.75$.

The $n$th term of an AP is given by the formula: $a_n = a + (n - 1)d$. [Formula for the $n$th term of an AP]

Step 2: Formulating the Equation

Substitute the given values into the formula:

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

Step 3: Solving for $n$

Subtract $5$ from both sides of the equation:

$20.75 - 5 = (n - 1) \times 1.75$

$15.75 = (n - 1) \times 1.75$

Divide both sides by $1.75$ to isolate the term $(n - 1)$:

$\frac{15.75}{1.75} = n - 1$

To simplify the division, multiply both numerator and denominator by $100$:

$\frac{1575}{175} = n - 1$

Perform the division:

$1575 \div 175 = 9$

$9 = n - 1$

Add $1$ to both sides to solve for $n$:

$n = 9 + 1$

$n = 10$

Step 4: Verification

If $n = 10$, then $a_{10} = 5 + (10 - 1) \times 1.75$

$a_{10} = 5 + 9 \times 1.75$

$a_{10} = 5 + 15.75$

$a_{10} = 20.75$

[Since the calculated $a_{10}$ matches the given $a_n$, the value of $n$ is correct.]

Final Answer: The value of $n$ is $10$.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.2


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


Chapters in CBSE - Class 10 Mathematics


Other Subjects in CBSE - Class 10

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