Question

6. Ashok deposits * 3,200 per month in a cumulative deposit account for 3 years at the rate
of 9% per annum. Find the maturity value of this account.
Mr Karna has a recurring deposit account in Dunial .​

Answer 1

3200+6oo+9600.......

sum of 36 terms.in Ap

s36=36/2{2x3200+ (36-1)3200}

= 18{64oo+35x3200}

= 18 (6400+ 112000)

= 18x1184oo=Rs2131200

Interest=2131200x9./l2oo=Rs15984

deposit =3200x36=115200+15984

= Rs. 131184. Ans

Answer 2

Concept Introduction: Maturity Value is the total amount payable at the end of the term

Given:

We have been Given: Deposit Amount is

$p = 3200$

time is,

$t = 3 \: years = 3 \times 12 = 36 \: months$

rate is,

$r = 9\% \: p.a.$

To Find:

We have to Find: The Maturity Value of the account.

Solution:

According to the problem, Total Simple Interest is,

$s.i. = \frac{p \times r \times n(n + 1)}{100 \times 12 \times 2}$

therefore putting the values,

$\frac{3200 \times 9 \times 36(36 + 1)}{2400} = \frac{3200 \times 9 \times 36 \times 37}{2400} = \frac{38361600}{2400} = 15984$

There the S.I of the amount is Rs.

$15984$

Now. Total Money deposited will be,

$money \: deposited \times no. \: of \: months$

therefore,

$3200 \times 36 = 115200$

Therefore now to calculate the Maturity Value,

$total \: deposited \: amount + total \: simple \: interest$

therefore, putting the values,

$115200 + 15984 = 131184$

Hence the Maturity Value of the Recurring Deposit Account is Rs.

$131184$

Final Answer: The Maturity Value of the Recurring Deposit Account is Rs.

$131184$

#SPJ2