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Mathematics

Mrs Goswami deposits ₹1000 per month in a recurring deposit account for 3 years at 8% interest per annum. Find the matured value.

Banking

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Answer

Given,

P = ₹1,000

n = 3 years = 3 x 12 = 36 months

r = 8%

I = P×n(n+1)2×12×r100P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100}

∴I=1000×36×372×12×8100I=1000×133224×0.08I=1000×55.5×0.08I=₹4,440\therefore I = 1000 \times \dfrac{36 \times 37}{2 \times 12} \times \dfrac{8}{100} \\[1em] I = 1000 \times \dfrac{1332}{24} \times 0.08 \\[1em] I = 1000 \times 55.5 \times 0.08 \\[1em] I = ₹4,440

Sum deposited = ₹1,000 x 36 = ₹36,000

Maturity value = Sum deposited + Interest = ₹36,000 + ₹4,440 = ₹40,440

Hence, the matured value is ₹40,440

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