Q43 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics
$\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\ldots . \text { to } 9 \text { terms }=$
- A$\frac{10}{99}$
- B$\frac{11}{108}$
- Correct answerC$\frac{1}{10}$
- D$\frac{1}{90}$
Solution
Let $S=\frac{1}{3 \cdot 6}+\frac{1}{6 \cdot 9}+\frac{1}{9 \cdot 12}+\ldots \ldots$ to 9 terms
$\begin{aligned} & S=\frac{1}{3}\left[\frac{6-3}{3 \cdot 6}+\frac{9-6}{6 \cdot 9}+\frac{12-9}{9 \cdot 12}+\ldots+\frac{30-27}{27 \cdot 30}\right] \\ & =\frac{1}{3}\left[\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\ldots . .+\frac{1}{27}-\frac{1}{30}\right] \\ & =\frac{1}{3}\left[\frac{1}{3}-\frac{1}{30}\right]=\frac{10-1}{90}=\frac{9}{90}=\frac{1}{10} .\end{aligned}$
Hence, the correct answer is option: 'C'.