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1999 Paper 3 Q1
D: 1700.0 B: 1500.0
roots of polynomials symmetric functions cubic equation geometric progression arithmetic progression Vieta's formulas necessary and sufficient condition

Consider the cubic equation \[ x^3-px^2+qx-r=0\;, \] where \(p\ne0\) and \(r\ne 0\).

  1. If the three roots can be written in the form \(ak^{-1}\), \(a\) and \(ak\) for some constants \(a\) and \(k\), show that one root is \(q/p\) and that \(q^3 -rp^3=0\;.\)
  2. If \(r=q^3/p^3\;\), show that \(q/p\) is a root and that the product of the other two roots is \((q/p)^2\). Deduce that the roots are in geometric progression.
  3. Find a necessary and sufficient condition involving \(p\), \(q\) and \(r\) for the roots to be in arithmetic progression.

Solution:

  1. If the roots are \(ak^{-1}, a, ak\) then we must have that \(p = a(k^{-1}+1+k)\), \(q = a^2(k^{-1}+k+1)\) and \(r = a^3\), therefore \(a = \frac{q}{p}\) (ie one of the roots is \(\frac q p\) and \(r = \left ( \frac{q}{p} \right)^3 \Rightarrow q^3 =rp^3 \Rightarrow q^3-rp^3 = 0\)
  2. Suppose \(r = q^3/p^3\) then \(\left (\frac{q}{p} \right)^3 - p\left (\frac{q}{p} \right)^2+q\left (\frac{q}{p} \right) - r = \frac{--pq^2+pq^2}{p^2} =0 \), therefore \(q/p\) is a root by the factor theorem. We must also have the product of the three roots is \(q^3/p^3\) but one of the roots is \(q/p\) therefore the product of the other two roots is \(q^2/p^2\), but the condition \(ac = b^2\) is precisely the condition that \(a,b,c\) is a geometric progression.
  3. If the three roots are \(a-d, a, a+d\) then \(p = 3a\), \(q = a^2-da+a^2+da+a^2-d^2 = 3a^2-d^2\), \(r = a(a^2-q^2)\), therefore \(\frac{p}{3}\left (q-\frac{2p^2}9 \right) = r\) Similarly, suppose \(\frac{p}{3}\) is a root, then the other two roots must sum to twice this and therefore they are in arithmetic progression. The condition \(\frac{p}{3}\) is a root is equivalent to: \(\frac{p^3}{27} - \frac{p^3}{9} + \frac{qp}{3} - r = 0\), ie exactly \(\frac{p}{3}\left (q-\frac{2p^2}9 \right) = r\), therefore this condition is both necessary and sufficient.

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1990 Paper 1 Q4
D: 1516.0 B: 1500.0
arithmetic progression sequences summation double arithmetic progression factorial progression difference method algebraic manipulation

The sequence \(a_{1},a_{2},\ldots,a_{n},\ldots\) forms an arithmetic progression. Establish a formula, involving \(n,\) \(a_{1}\) and \(a_{2}\) for the sum \(a_{1}+a_{2}+\cdots+a_{n}\) of the first \(n\) terms. A sequence \(b_{1},b_{2},\ldots,b_{n},\ldots\) is called a double arithmetic progression if the sequence of differences \[ b_{2}-b_{1},b_{3}-b_{2},\ldots,b_{n+1}-b_{n},\ldots \] is an arithmetic progression. Establish a formula, involving \(n,b_{1},b_{2}\) and \(b_{3}\), for the sum \(b_{1}+b_{2}+b_{3}+\cdots+b_{n}\) of the first \(n\) terms of such a progression. A sequence \(c_{1},c_{2},\ldots,c_{n},\ldots\) is called a factorial progression if \(c_{n+1}-c_{n}=n!d\) for some non-zero \(d\) and every \(n\geqslant1\). Suppose \(1,b_{2},b_{3},\ldots\) is a double arithmetic progression, and also that \(b_{2},b_{4},b_{6}\) and \(220\) are the first four terms in a factorial progression. Find the sum \(1+b_{2}+b_{3}+\cdots+b_{n}.\)

Solution: Since the common difference is \(a_2 - a_1\) we can find that \(a_n = a_1 + (n-1)(a_2-a_1)\), then \begin{align*} && a_1 &&+&& a_2 &&+&& \cdots &&+&& (a_1 + (n-2)(a_2 - a_1) && + && (a_1 + (n-1) (a_2 - a_1)) \\ + && (a_1 + (n-1) (a_2 - a_1))&&+&& (a_1 + (n-2)(a_2 - a_1)&&+&& \cdots &&+&& a_2 && + && a_1 \\ \hline \\ = && 2a_1 + (n-1)(a_2 - a_1) && + && 2a_1 + (n-1)(a_2 - a_1) && + && \cdots && + 2a_1 + (n-1)(a_2 - a_1) && + 2a_1 + (n-1)(a_2 - a_1) \\ = && n(2a_1 + (n-1) (a_2 - a_1)) \end{align*} Therefore the sum is \(a_1 n + \frac{n(n-1)}{2} (a_2 - a_1)\). Since \(b_n - b_1 = (b_n - b_{n-1}) + (b_{n-1} - b_{n-2}) + \cdots (b_2 - b_1)\), \(b_n - b_1 = a_1 (n-1) + \frac{(n-1)(n-2)}{2}(a_2 - a_1) = (b_2-b_1)(n-1) + \frac{(n-1)(n-2)}{2}(b_3 -2b_2 +b_1)\). So \(b_n = b_1 + (b_2 - b_1)(n-1) + \frac{(n-1)(n-2)}{2} (b_3 - 2b_2 + b_1)\). In particular \begin{align*} \sum_{i=1}^n b_i &= \sum_{i=1}^n \l b_1 + (b_2 - b_1)(n-1) + \frac{(n-1)(n-2)}{2} (b_3 - 2b_2 + b_1)\r \\ &= nb_1 + (b_2-b_1) \frac{n(n-1)}{2} + \frac{n(n-1)(n-2)}{6}(b_3-2b_2+b_1) \end{align*} Let \(b_2 - b_1 = x\) and \(b_3 - 2b_2+b_1 = y\), then \begin{align*} b_4 - b_2 &= d &= &2x + 3y \\ b_6-b_4 &= 2d &=& 2x +(10-3)y \\ &&=&2x + 7y \\ 220-b_6&=6d &=& 220-(1 + 5x + 10y) \\ \end{align*} \begin{align*} && 4x + 6y &= 2x + 7y \\ && 6x+21y &= 219-5x-10y \\ \Rightarrow && 2x - y &= 0 \\ && 11x + 31y &= 219 \\ \Rightarrow && x &= 3 \\ && y &= 6 \end{align*} Therefore the final sum is \begin{align*} n + 3 \frac{n(n-1)}{2} + 6 \frac{n(n-1)(n-2)}{6} &= n^3-\frac32n^2+\frac32n \end{align*}

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