Postgraduate Mathematics II task 87: An arithmetic sequence has first term 2 and common difference 4. Enter its 14th term.
aₙ=a₁+(n−1)d=2+(14−1)×4=54.
Practice China questions with answers and explanations.
aₙ=a₁+(n−1)d=2+(14−1)×4=54.
An antiderivative is 1/2 x²+5x. At x=2, the value is 12.0.
The dot product is 1×4+6×3=22.
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.
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Use the circle-area formula A = πr². Substituting r = 5 gives A = π × 5² = 25π.
Percentage=(8/33)×100=24.24.
Subtract 11 from both sides and divide by 4: x = (15-11)/4 = 1.
By Vieta’s formula, the sum of the roots equals the coefficient 16.
aₙ=a₁+(n−1)d=10+(13−1)×3=46.
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The sum is n(n+1)/2=21, so the mean is 21/6.
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There are 6 favourable outcomes out of 11 equally likely outcomes, so P(red)=6/11.