Postgraduate Mathematics I task 15: An arithmetic sequence has first term 5 and common difference 4. Enter its 14th term.
aₙ=a₁+(n−1)d=5+(14−1)×4=57.
Practice China questions with answers and explanations.
aₙ=a₁+(n−1)d=5+(14−1)×4=57.
Rewrite the value using base 2: 4 = 2^2. Matching exponents gives x = 2.
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Use the circle-area formula A = πr². Substituting r = 6 gives A = π × 6² = 36π.
Subtract 5 from both sides and divide by 8: x = (85-5)/8 = 10.
By Vieta’s formula, the sum of the roots equals the coefficient 13.
aₙ=a₁+(n−1)d=4+(13−1)×3=40.
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The sum is n(n+1)/2=55, so the mean is 55/10.
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There are 2 favourable outcomes out of 7 equally likely outcomes, so P(red)=2/7.
The coordinate differences are 3 and 4, so the distance is √(3²+4²)=5.
f′(x)=3x²+16x. Substituting x=1 gives 19.
An antiderivative is 1/2 x²+1x. At x=2, the value is 4.0.
The dot product is 1×4+2×3=10.