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Mathematics

A certain sum of money (₹ P) is lent for 3123\dfrac{1}{2} years at r% C.I. compounded half yearly. The interest accrued will be:

  1. P(1+r100)72PP\Big(1 + \dfrac{r}{100}\Big)^{\dfrac{7}{2}} - P

  2. P(1+r100)3×(1+r2×100)1PP\Big(1 + \dfrac{r}{100}\Big)^3 \times \Big(1 + \dfrac{r}{2 \times 100}\Big)^1 - P

  3. P(1+r2×100)72PP\Big(1 + \dfrac{r}{2 \times 100}\Big)^{\dfrac{7}{2}} - P

  4. P(1+r2×100)7PP\Big(1 + \dfrac{r}{2 \times 100}\Big)^7 - P

Compound Interest

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Answer

Given, the principal amount = ₹ P.

The time period = 312=723\dfrac{1}{2} = \dfrac{7}{2} years.

The annual interest rate = r% .

Interest is compounded half-yearly.

The formula for C.I. when compounded half-yearly is I

= P (1+r2×100)n×2P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2} - P

Substituting the values, we get :

I=P(1+r2×100)72×2P=P(1+r2×100)7PI = P \Big(1 + \dfrac{r}{2 \times 100}\Big)^{\dfrac{7}{2} \times 2} - P\\[1em] = P \Big(1 + \dfrac{r}{2 \times 100}\Big)^7 - P

Hence, option 4 is the correct option.

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