Linked Progressions | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Three consecutive terms give an equation

For three numbers $A,B,C$ in an AP, their differences agree:

\begin{gathered}B-A=C-B\\2B=A+C\end{gathered}

For three non-zero numbers in a GP, their ratios agree:

\frac BA=\frac CB\quad\Rightarrow\quad B^2=AC

Use the numbers in the order given. After solving, check the original differences or ratios. If a term is zero, check multiplication directly: $0,0,5$ satisfies $B^2=AC$ but cannot be generated by one fixed multiplier.

Worked example

The numbers $x,4,x+6$ are consecutive GP terms. Find $x$ .

\begin{gathered}16=x(x+6)\\x^2+6x-16=0\end{gathered}

The polynomial solver gives 2 and −8, so write $(x-2)(x+8)=0$ . Both work: $2,4,8$ has ratio 2; $-8,4,-2$ has ratio $-1/2$ .

Keep the two progressions separate

Worked example

An AP begins $x,y,x^2$ . A GP begins $x,x^2,y$ . Given $x<0$ , find $x,y$ and the GP sum to infinity.

The AP gives $2y=x+x^2$ . The GP gives $x^4=xy$ . Since $x<0$ , divide by non-zero $x$ to obtain $y=x^3$ .

\begin{gathered}2x^3=x+x^2\\2x^2-x-1=0\end{gathered}

The solver gives $x=-1/2,1$ , supporting $(2x+1)(x-1)=0$ . The negative condition selects $x=-1/2$ , then $y=-1/8$ .

The GP has its own first term $-1/2$ and ratio $(1/4)/(-1/2)=-1/2$ . Thus $S_\infty=(-1/2)/(1+1/2)=-1/3$ . The AP difference is instead $3/8$ .

If two sums are equal, write one sum formula for each progression and equate them. If an AP supplies spaced GP terms, translate their AP positions first; they are consecutive in the GP, not necessarily in the AP.

Now you try

Your turnIndependent practice

(a) A sequence starts $10,10+d,10+2d,\ldots$ , with constant difference $d>0$ . Its first, third and seventh terms form a GP. Find $d$ and the ratio of that GP.

(b) An AP begins $1,\cos^2\theta,\ldots$ . Show that $S_5=5-10\sin^2\theta$ .

Show worked answer

(a) The selected terms are $10,10+2d,10+6d$ .

(10+2d)^2=10(10+6d)

4d(d-5)=0

$d>0$ selects 5. The terms 10, 20, 40 have ratio 2.

(b) $d=\cos^2\theta-1=-\sin^2\theta$ . Therefore

\begin{aligned}S_5&=\frac52[2-4\sin^2\theta]\\&=5-10\sin^2\theta\end{aligned}

Three consecutive terms give an equation

For three numbers $A,B,C$ in an AP, their differences agree:

\begin{gathered}B-A=C-B\\2B=A+C\end{gathered}

For three non-zero numbers in a GP, their ratios agree:

\frac BA=\frac CB\quad\Rightarrow\quad B^2=AC

Worked example

The numbers $x,4,x+6$ are consecutive GP terms. Find $x$ .

\begin{gathered}16=x(x+6)\\x^2+6x-16=0\end{gathered}

The polynomial solver gives 2 and −8, so write $(x-2)(x+8)=0$ . Both work: $2,4,8$ has ratio 2; $-8,4,-2$ has ratio $-1/2$ .

Keep the two progressions separate

Worked example

An AP begins $x,y,x^2$ . A GP begins $x,x^2,y$ . Given $x<0$ , find $x,y$ and the GP sum to infinity.

The AP gives $2y=x+x^2$ . The GP gives $x^4=xy$ . Since $x<0$ , divide by non-zero $x$ to obtain $y=x^3$ .

\begin{gathered}2x^3=x+x^2\\2x^2-x-1=0\end{gathered}

The solver gives $x=-1/2,1$ , supporting $(2x+1)(x-1)=0$ . The negative condition selects $x=-1/2$ , then $y=-1/8$ .

The GP has its own first term $-1/2$ and ratio $(1/4)/(-1/2)=-1/2$ . Thus $S_\infty=(-1/2)/(1+1/2)=-1/3$ . The AP difference is instead $3/8$ .

Now you try

Your turnIndependent practice

(a) A sequence starts $10,10+d,10+2d,\ldots$ , with constant difference $d>0$ . Its first, third and seventh terms form a GP. Find $d$ and the ratio of that GP.

(b) An AP begins $1,\cos^2\theta,\ldots$ . Show that $S_5=5-10\sin^2\theta$ .

Show worked answer

(a) The selected terms are $10,10+2d,10+6d$ .

(10+2d)^2=10(10+6d)

4d(d-5)=0

$d>0$ selects 5. The terms 10, 20, 40 have ratio 2.

(b) $d=\cos^2\theta-1=-\sin^2\theta$ . Therefore

\begin{aligned}S_5&=\frac52[2-4\sin^2\theta]\\&=5-10\sin^2\theta\end{aligned}