Senior-secondary mathematics task 159: An arithmetic sequence has first term 7 and common difference 3. Enter its 7th term.
aₙ=a₁+(n−1)d=7+(7−1)×3=25.
Algebra, arithmetic, reasoning.
aₙ=a₁+(n−1)d=7+(7−1)×3=25.
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The sum is n(n+1)/2=10, so the mean is 10/4.
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There are 6 favourable outcomes out of 11 equally likely outcomes, so P(red)=6/11.
The coordinate differences are 3 and 4, so the distance is √(3²+4²)=5.
The coordinate differences are 3 and 4, so the distance is √(3²+4²)=5.
f′(x)=12x²+16x. Substituting x=2 gives 80.
An antiderivative is 2/2 x²+4x. At x=3, the value is 21.0.
The dot product is 2×3+5×2=16.
Rewrite the value using base 3: 27 = 3^3. Matching exponents gives x = 3.
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Use the circle-area formula A = πr². Substituting r = 7 gives A = π × 7² = 49π.
Percentage=(8/33)×100=24.24.
Subtract 8 from both sides and divide by 2: x = (16-8)/2 = 4.