Karnataka SSLC 10th Maths Question Paper 2022

Given AP: 7, 11, 15, ... First term $a = 7$, common difference $d = 11 - 7 = 4$, and number of terms $n = 16$. The sum of $n$ terms is given by: $S_n = \frac{n}{2} [2a + (n-1)d]$ Substituting the values: $S_{16} = \frac{16}{2} [2(7) + (16-1)4]$ $S_{16} = 8 [14 + (15 \times 4)]$ $S_{16} = 8 [14 + 60]$ $S_{16} = 8 \times 74 = 592$ The sum of the first 16 terms is 592.

Given AP: 3, 6, 9, ... Here, $a = 3$, $d = 3$, and $S_n = 165$. Using the sum formula $S_n = \frac{n}{2} [2a + (n-1)d]$: $165 = \frac{n}{2} [2(3) + (n-1)3]$ $330 = n [6 + 3n - 3]$ $330 = n [3n + 3]$ $330 = 3n^2 + 3n$ Divide by 3: $n^2 + n - 110 = 0$ $(n + 11)(n - 10) = 0$ $n = -11$ (not possible as terms cannot be negative) or $n = 10$. Therefore, 10 terms must be added.

For the quadratic equation $4x^2 - 12x + 9 = 0$: Here, $a = 4, b = -12, c = 9$. Discriminant $D = b^2 - 4ac$ $D = (-12)^2 - 4(4)(9)$ $D = 144 - 144 = 0$ Since the discriminant is 0, the roots of the equation are real and equal.

For $x^2 - 3x + 1 = 0$, $a = 1, b = -3, c = 1$. The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)}$ $x = \frac{3 \pm \sqrt{9 - 4}}{2}$ $x = \frac{3 \pm \sqrt{5}}{2}$ So, the roots are $x = \frac{3 + \sqrt{5}}{2}$ and $x = \frac{3 - \sqrt{5}}{2}$.

In triangle ABC, for angle $\alpha$ (at vertex C): Opposite side = $AB = 24$ cm Adjacent side = $BC = 7$ cm Hypotenuse = $AC = 25$ cm $\sin \alpha = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$ $\cos \alpha = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$