Govt Exams
By Fermat's Little Theorem, 2⁶ ≡ 1 (mod 7). So 2⁵⁰ = 2⁴⁸ × 2² = (2⁶)⁸ × 4 ≡ 1⁸ × 4 ≡ 4 (mod 7).
Let y = 5^x. Then y² - 6y + 5 = 0, so (y-5)(y-1) = 0. Thus y = 5 or y = 1, giving x = 1 or x = 0
nC2 = n(n-1)/2 = 21 means n(n-1) = 42. For n=7: 7×6 = 42. So n = 7
We know sec²θ - tan²θ = 1. So (sec θ - tan θ)(sec θ + tan θ) = 1. Given sec θ - tan θ = 1/5, we get (1/5)(sec θ + tan θ) = 1, so sec θ + tan θ = 5
sin 3θ = cos θ = sin(90° - θ). So 3θ = 90° - θ → 4θ = 90° → θ = 22.5°.
(tan θ + cot θ)² = tan²θ + 2 + cot²θ = 16. So tan²θ + cot²θ = 14.
2sin²θ - sin θ - 1 = 0. Factoring: (2sinθ + 1)(sinθ - 1) = 0. Since θ is acute, sin θ = 1.
S₅ = a(r⁵ - 1)/(r - 1) = 2(32 - 1)/(2 - 1) = 2×31 = 62
Mode = 3×Median - 2×Mean, 23 = 3×Median - 50, Median = 24.33. Actually: Median = (Mode + 2×Mean)/3 = (23 + 50)/3 = 24.33, checking: mode≈3med-2mean gives med≈24.67