Assertion (A): Pawandeep opened a recurring deposit account in a bank for a period of 2 years. If the bank pays interest at the rate of 6% p.a. and the monthly instalment is ₹1,000, then the maturity amount is ₹25,000.

Reason (R): For a recurring deposit account, we compute the interest using the following formula:

$S.I. = P \times \dfrac{n(n+1)}{ 12} \times \dfrac{R}{100}\\[1em]$

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false.

Both A and R are false.

Reason

According to Assertion:

The maturity amount is ₹25,000.

Given,

P = ₹1,000

n = 2 years = 24 months

r = 6%

I = $P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100}$

$\therefore I = 1000\times \dfrac{24\times 25}{2 \times 12} \times \dfrac{6}{100} \\[1em] I = 1000 \times \dfrac{600}{24} \times 0.06\\[1em] I = 1000 \times 25 \times 0.06 \\[1em] I = ₹1500$

Sum deposited = ₹1,000 x 24 = ₹24,000

Maturity value = Sum deposited + Interest = ₹24,000 + ₹1,500 = ₹25,500

The given Maturity amount = ₹25,500

So, Assertion(A) is false.

For a recurring deposit account, we compute the interest using the following formula:

I = $P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100}$

According to Reason:

For a recurring deposit account, we compute the interest using the following formula:

$I = P \times \dfrac{n(n+1)}{ 12} \times \dfrac{R}{100}\\[1em]$

So, Reason(R) is false .

Hence, Option 4 is the correct option.