Selected Geometric Series | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Give the selected series its own inputs

A contains selected terms of a sequence. If the positions are equally spaced in a GP, the selected terms also form a GP.

Original	Selected	Value
2	1	ar
5	2	ar⁴
8	3	ar⁷

r × r × r = r³

between selected terms

Each gap contains three multiplications by r.

For $u_2,u_5,u_8,\ldots$ , the first selected term is $ar$ . Each gap contains three original multiplications, so the new ratio is $r^3$ . For the first $m$ selected terms, the count is $m$ , not the last original position.

Worked example

A GP has $a=3$ and $r=2$ . Add $u_2+u_4+u_6$ .

The selected terms start at 6 and have ratio 4. There are three terms:

S=\frac{6(4^3-1)}{4-1}=126

Check: $6+24+96=126$ .

The terms remaining after the first n

The remaining part, or tail, starts at $u_{n+1}=ar^n$ . It has the same ratio as the original GP. When $|r|<1$ :

S_\infty-S_n=\frac{ar^n}{1-r}

For $8-4+2-1+\cdots$ , the tail after the first three terms starts at −1. Its sum is $-1/(1+1/2)=-2/3$ . A remaining sum can be negative. If a question asks for the size of the error, use its absolute value.

Now you try

Your turn9709/12 Feb/Mar 2024 Q8(b) [4 marks]

A GP has first term 24 and ratio $1/2$ . Find its sum to infinity $S$ , and the sum $S_E$ of its even-numbered terms.

Show worked answer

$S=24/(1-1/2)=48$ . The even terms start at 12 and have ratio $(1/2)^2=1/4$ , so $S_E=12/(1-1/4)=16$ .

Answer

S=48,\qquad S_E=16

Your turnIndependent practice

For $6-2+\tfrac23-\cdots$ , find the sum remaining after the first four terms.

Show worked answer

The next term is $6(-1/3)^4=2/27$ . Divide by $1-(-1/3)=4/3$ to get $1/18$ . This tail is positive because its first term is positive.

Give the selected series its own inputs

A contains selected terms of a sequence. If the positions are equally spaced in a GP, the selected terms also form a GP.

Original	Selected	Value
2	1	ar
5	2	ar⁴
8	3	ar⁷

r × r × r = r³

between selected terms

Each gap contains three multiplications by r.

Worked example

A GP has $a=3$ and $r=2$ . Add $u_2+u_4+u_6$ .

The selected terms start at 6 and have ratio 4. There are three terms:

S=\frac{6(4^3-1)}{4-1}=126

Check: $6+24+96=126$ .

The terms remaining after the first n

The remaining part, or tail, starts at $u_{n+1}=ar^n$ . It has the same ratio as the original GP. When $|r|<1$ :

S_\infty-S_n=\frac{ar^n}{1-r}

Now you try

Your turn9709/12 Feb/Mar 2024 Q8(b) [4 marks]

A GP has first term 24 and ratio $1/2$ . Find its sum to infinity $S$ , and the sum $S_E$ of its even-numbered terms.

Show worked answer

$S=24/(1-1/2)=48$ . The even terms start at 12 and have ratio $(1/2)^2=1/4$ , so $S_E=12/(1-1/4)=16$ .

Answer

S=48,\qquad S_E=16

Your turnIndependent practice

For $6-2+\tfrac23-\cdots$ , find the sum remaining after the first four terms.

Show worked answer

The next term is $6(-1/3)^4=2/27$ . Divide by $1-(-1/3)=4/3$ to get $1/18$ . This tail is positive because its first term is positive.