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    PracticeAQA GCSEAQA GCSE Maths Higher Tier Paper 2 CalculatorQuestion 22
    Hard3 marksStructured
    AlgebraHigheralgebracompleting the squaresequences

    AQA GCSE · Question 22 · Algebra

    The nth term of a sequence is n² - 30n + 236
    <br>
    By completing the square, show that all the terms of the sequence have two or more digits.

    How to approach this question

    1. Take the expression for the nth term: n² - 30n + 236. 2. Complete the square for the n² and n terms. To do this, halve the coefficient of n (-30) to get -15. Write this as (n - 15)². 3. You must then subtract the square of this number (-15)² to balance the expression. 4. Don't forget to include the original constant term (+236). 5. Simplify the constant terms to get the final completed square form: (n - 15)² + k. 6. Consider the minimum possible value of this expression. The squared bracket part, (n - 15)², can never be negative. What is its minimum value? 7. Use this to find the minimum value of the whole expression, and explain why this shows all terms have at least two digits.

    Full Answer

    **Step 1: Complete the square for the expression n² - 30n + 236.** To complete the square for n² - 30n, we take half of the coefficient of n (-30), which is -15, and square it. The expression becomes: (n - 15)² - (-15)² + 236 **Step 2: Simplify the expression.** = (n - 15)² - 225 + 236 = (n - 15)² + 11 **Step 3: Analyse the completed square form to find the minimum value.** The term (n - 15)² is a squared number, so its minimum possible value is 0. This occurs when n = 15. For any integer value of n, (n - 15)² ≥ 0. Therefore, the minimum value of the entire expression (n - 15)² + 11 will occur when (n - 15)² is at its minimum. Minimum value = 0 + 11 = 11. **Step 4: Conclude based on the minimum value.** The minimum value of any term in the sequence is 11. Since 11 is a two-digit number, and all other terms will be greater than 11, all terms of the sequence must have two or more digits. For example, if n=1, the term is (1-15)² + 11 = (-14)² + 11 = 196 + 11 = 207. If n=15, the term is (15-15)² + 11 = 0 + 11 = 11.
    We are given the nth term as Tₙ = n² - 30n + 236. We need to complete the square for this expression. 1. **Halve the coefficient of n:** The coefficient of n is -30. Half of -30 is -15. 2. **Write the squared bracket:** This gives us (n - 15)². 3. **Expand this bracket to see what we have:** (n - 15)² = n² - 30n + 225. 4. **Adjust the constant:** Our original expression has +236, but the bracket gives +225. To get from 225 to 236, we need to add 11. So, n² - 30n + 236 = (n² - 30n + 225) + 11 = (n - 15)² + 11. The nth term of the sequence is given by (n - 15)² + 11. Now we need to show that all terms have two or more digits. We can do this by finding the minimum value of the expression. - The term (n - 15)² is a square, so its value is always greater than or equal to 0 for any integer n. - The minimum value of (n - 15)² is 0, which occurs when n = 15. - Therefore, the minimum value of the entire expression is 0 + 11 = 11. Since the smallest possible value of any term in the sequence is 11, and all other terms will be larger, every term in the sequence is a positive integer greater than or equal to 11. All such numbers have two or more digits.

    Common mistakes

    ✗ Errors in the process of completing the square, especially with the constant term. ✗ Successfully completing the square but failing to provide a clear explanation of why this shows the minimum value is 11. ✗ Thinking the minimum value is just the constant term (11) without considering the squared part.
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