Question

Mrs Goswami deposits rs 1000 every month in a recurring deposit account for 3 years at8% interest per annum. find the maturity value​

Answer 1

Solution :

$\underline{\bf{Given\::}}}}$

• Installment, (P) = Rs.1000
• Rate, (R) = 8%
• Time, (n) = 3 years [∴ Every month = 12 × 3 = 36 months ]

$\underline{\bf{To\:find\::}}}}$

The maturity value.

$\underline{\bf{Explanation\::}}}}$

Using formula of the deposits Interest :

$\longrightarrow\sf{Interest=\dfrac{n(n+1)}{2} \times \dfrac{Principal\times Rate}{100\times 12} }\\\\\\\longrightarrow\sf{Interest=\dfrac{36(36+1)}{2} \times\dfrac{10\cancel{00}\times 8}{\cancel{100} \times 12} }\\\\\\\longrightarrow\sf{Interest=\dfrac{36\times 37}{2} \times \dfrac{80}{12} }\\\\\\\longrightarrow\sf{Interest=\dfrac{36\times 37\times 80}{24} }\\\\\\\longrightarrow\sf{Interest=\cancel{\dfrac{106560}{24} }}\\\\\longrightarrow\bf{Interest=Rs.4440}$

Thus;

$\longrightarrow\tt{Maturity\:value=Pn+Interest}\\\\\longrightarrow\tt{Maturity\:value=1000\times 36+4440}\\\\\longrightarrow\tt{Maturity\:value=Rs.(36000+4440)}\\\\\longrightarrow\bf{Maturity\:value=Rs.40440}$