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Question 9
An arithmetic sequence has first term a and common difference d. The sum of the first 36 terms of the sequence is equal to the square of the sum of the first 6 term... show full transcript
Step 1
Answer
To find the sixth term of the sequence, we use the expression for the nth term:
For the sixth term, where :
We know from the problem statement that :
Rearranging this gives:
Now, substituting this expression for into the quadratic equation we derived in part (a):
Expanding and simplifying gives:
Eventually, setting will yield:
Verifying with the quadratic equation shows that this is the smallest achievable value for . Thus, the solution yields: \smallest possible value of a = -55.
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1.1 Proof
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1.2 Proof by Contradiction
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2.1 Laws of Indices & Surds
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2.2 Quadratics
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2.3 Simultaneous Equations
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2.4 Inequalities
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2.5 Polynomials
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2.6 Rational Expressions
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2.7 Graphs of Functions
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2.8 Functions
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2.9 Transformations of Functions
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2.10 Combinations of Transformations
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2.11 Partial Fractions
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2.12 Modelling with Functions
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2.13 Further Modelling with Functions
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3.1 Equation of a Straight Line
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3.2 Circles
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4.1 Binomial Expansion
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4.2 General Binomial Expansion
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4.3 Arithmetic Sequences & Series
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4.4 Geometric Sequences & Series
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4.5 Sequences & Series
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4.6 Modelling with Sequences & Series
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5.1 Basic Trigonometry
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5.2 Trigonometric Functions
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5.3 Trigonometric Equations
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5.4 Radian Measure
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5.5 Reciprocal & Inverse Trigonometric Functions
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5.6 Compound & Double Angle Formulae
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5.7 Further Trigonometric Equations
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5.8 Trigonometric Proof
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5.9 Modelling with Trigonometric Functions
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6.1 Exponential & Logarithms
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6.2 Laws of Logarithms
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6.3 Modelling with Exponentials & Logarithms
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7.1 Differentiation
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7.2 Applications of Differentiation
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7.3 Further Differentiation
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7.4 Further Applications of Differentiation
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7.5 Implicit Differentiation
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8.1 Integration
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8.2 Further Integration
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8.3 Differential Equations
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9.1 Parametric Equations
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10.1 Solving Equations
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10.2 Modelling involving Numerical Methods
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11.1 Vectors in 2 Dimensions
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11.2 Vectors in 3 Dimensions
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