Discussion

Explanation:

a₁ = $\frac{1}{2\times 5}$ = $\frac{1}{3}\left(\frac{3}{2\times 5}\right)$ = $\frac{1}{3}\left(\frac{5-2}{2\times 5}\right)$ = $\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}\right)$

Similarly, 
a₂ = $\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}\right)$

a₃ = $\frac{1}{3}\left(\frac{1}{8}-\frac{1}{11}\right)$ and so on till

a₁₀₀ = $\frac{1}{3}\left(\frac{1}{299}-\frac{1}{302}\right)$ [First term of a₁₀₀ = 2 + (100 - 1) × 3 = 299]

Now if we add a₁, a₂, a₃, ...., a₁₀₀

a₁ + a₂ + a₃ + ... + a₁₀₀ = $\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}\right)$ + $\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}\right)$ + $\frac{1}{3}\left(\frac{1}{8}-\frac{1}{11}\right)$ + ... + $\frac{1}{3}\left(\frac{1}{299}-\frac{1}{302}\right)$

= $\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{299}-\frac{1}{302}\right)$

Terms $\frac{1}{5}$, $\frac{1}{8}$, ... $\frac{1}{296}$, $\frac{1}{299}$ will all cancel out

⇒ a₁ + a₂ + a₃ + ... + a₁₀₀ = $\frac{1}{3}\left(\frac{1}{2}-\frac{1}{302}\right)$

⇒ a₁ + a₂ + a₃ + ... + a₁₀₀ = $\frac{1}{3}\left(\frac{1}{2}-\frac{1}{302}\right)$

So a₁ + a₂ + a₃ + ... + a₁₀₀ = $\frac{150}{906}$ = $\frac{25}{151}$

Hence, option (a).

» Your doubt will be displayed only after approval.