Anil invests some money at a fixed rate of interest, compounded annually. If the interests accrued during the second and third year are $₹ 806.25$ and $₹ 866.72,$ respectively, the interest accrued, in INR, during the fourth year is nearest to

In compound interest problems, there's a beautiful pattern that emerges when we look at the interest earned in consecutive years.

The interest earned in consecutive years forms a geometric progression with a common ratio of $(1 + \tfrac{r}{100})$, where $r$ is the rate of interest.

This means:

$\tfrac{\text{Interest in 3rd year}}{\text{Interest in 2nd year}} = \tfrac{\text{Interest in 4th year}}{\text{Interest in 3rd year}} = (1 + \tfrac{r}{100})$

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Let's think about what happens each year:

2nd year interest = Interest earned on the amount present at the start of 2nd year

3rd year interest = Interest earned on the amount present at the start of 3rd year

4th year interest = Interest earned on the amount present at the start of 4th year

Since the amount grows by the same factor $(1 + \tfrac{r}{100})$ each year, the interest earned also grows by the same factor.

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Given:

Interest in 2nd year = ₹806.25

Interest in 3rd year = ₹866.72

Using our pattern:

$\tfrac{\text{Interest in 4th year}}{\text{Interest in 3rd year}} = \tfrac{\text{Interest in 3rd year}}{\text{Interest in 2nd year}}$

Cross-multiplying:

$\text{Interest in 4th year} = \tfrac{(\text{Interest in 3rd year})^2}{\text{Interest in 2nd year}}$

$\text{Interest in 4th year} = \tfrac{(866.72)^2}{806.25}$

This can be written as:

$\text{Interest in 4th year} = \tfrac{866.72}{806.25} \times 866.72$

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$\tfrac{866.72}{806.25} = 1.075$

$1.075 \times 866.72 = 931.724$

Therefore, the interest accrued in the fourth year is approximately ₹931.72

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This approach is much faster than finding the principal and rate separately because it uses the inherent geometric progression property of compound interest. The ratio between consecutive year interests remains constant, making it a powerful shortcut for these types of problems.

In compound interest, consecutive year interests always have the same ratio, which equals $(1 + \tfrac{r}{100})$.