TS EAMCET · Maths · Determinants
If \(\Delta_{\mathrm{r}}=\left|\begin{array}{cc}\frac{1}{3 \mathrm{r}-2} & \frac{2}{3 \mathrm{r}-5} \\ 0 & \frac{3}{3 \mathrm{r}+1}\end{array}\right|\), then \(\sum_{\mathrm{r}=1}^{33} \Delta_{\mathrm{r}}=\)
- A \(0.99\)
- B \(0.33\)
- C \(0.66\)
- D \(0.55\)
Answer & Solution
Correct Answer
(A) \(0.99\)
Step-by-step Solution
Detailed explanation
\(\Delta_{\mathrm{r}}=\frac{1}{3 \mathrm{r}-2} \cdot \frac{3}{3 \mathrm{r}+1} - 0 = \frac{3}{(3 \mathrm{r}-2)(3 \mathrm{r}+1)}\) \(\Delta_{\mathrm{r}}=\frac{1}{3 \mathrm{r}-2} - \frac{1}{3 \mathrm{r}+1}\)…
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