CBSE 10th Maths Question Paper 2022 (Set 1)

Let the first term be $a$ and common difference be $d$. Then $a_n = a + (n-1)d$. Given $\frac{a_3}{a_2} = \frac{7}{4} \implies \frac{a+2d}{a+d} = \frac{7}{4}$. Cross-multiplying: $4a + 8d = 7a + 7d \implies 3a = d$. Now, $\frac{a_8}{a_6} = \frac{a+7d}{a+5d}$. Substituting $d = 3a$: $\frac{a+7(3a)}{a+5(3a)} = \frac{a+21a}{a+15a} = \frac{22a}{16a} = \frac{11}{8}$.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ for $2x^2 - 22x + 1 = 0$: $a=2, b=-22, c=1$. $x = \frac{22 \pm \sqrt{(-22)^2 - 4(2)(1)}}{2(2)} = \frac{22 \pm \sqrt{484 - 8}}{4} = \frac{22 \pm \sqrt{476}}{4}$. Since $\sqrt{476} = \sqrt{4 \times 119} = 2\sqrt{119}$, we get $x = \frac{22 \pm 2\sqrt{119}}{4} = \frac{11 \pm \sqrt{119}}{2}$.

For real and equal roots, the discriminant $D = b^2 - 4ac$ must be $0$. Here $a=1, b=5k, c=16$. Thus, $(5k)^2 - 4(1)(16) = 0 \implies 25k^2 - 64 = 0 \implies 25k^2 = 64 \implies k^2 = \frac{64}{25}$. Therefore, $k = \pm \frac{8}{5}$.

The A.P. is $293, 285, 277, ..., 53$. Here $a = 293$ and $d = 285 - 293 = -8$. Using $a_n = a + (n-1)d$, set $a_n = 53$: $53 = 293 + (n-1)(-8) \implies 53 - 293 = -8(n-1) \implies -240 = -8(n-1) \implies n-1 = 30 \implies n = 31$. The number of terms is 31.

Cumulative Frequency (CF) is calculated by adding frequencies: $CF_1 = 5$. $c = 5 + 7 = 12$. $18 = c + a \implies 18 = 12 + a \implies a = 6$. $d = 18 + 5 = 23$. $30 = d + b \implies 30 = 23 + b \implies b = 7$. Thus, $a=6, c=12, d=23, b=7$.