Selecting and Counting AP Terms | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Turn value bounds into term positions

“Terms between 20 and 50” refers to their values, not their positions. Find the first and last allowed terms before adding them.

Worked example

Add the terms of $5,11,17,\ldots$ strictly between 20 and 50.

Here $u_n=6n-1$ . “Strictly between” excludes both boundaries:

\begin{gathered}20<6n-1<50\\3.5<n<8.5\end{gathered}

The integer positions are 4 through 8, giving first term 23 and last term 47. Count both ends: $8-4+1=5$ terms.

S=\frac52(23+47)=175

For inclusive bounds, use $\le$ . If the difference is negative, dividing an inequality by it reverses the sign. A short calculator table of $u_n$ can check the two endpoint positions.

When the sum gives the number of terms

Worked example

How many terms of $2,5,8,\ldots$ sum to 155?

Use the sum formula because 155 is a total:

\frac n2[4+3(n-1)]=155

3n^2+n-310=0

The polynomial solver gives 10 and $-31/3$ . These convenient roots give the working $(n-10)(3n+31)=0$ . Only $n=10$ is a positive integer.

For a least number of terms exceeding a target, check the two neighbouring integers in the original sum. Here $S_{10}=155$ and $S_{11}=187$ , so 11 terms are needed to exceed 155. This comparison works because all these terms are positive and the sums increase.

Common mistake

Do not round a non-integer solution of an exact sum equation to make it a term count. It may mean no such count exists.

Now you try

Your turnIndependent practice

(a) Add all multiples of 7 strictly between 50 and 100.

(b) Find the least number of terms of $4,8,12,\ldots$ whose sum exceeds 220.

Show worked answer

(a) The terms run from 56 to 98, seven terms:

S=\frac72(56+98)=539

(b) $S_n=2n(n+1)$ . The sums increase since every term is positive. $S_{10}=220$ and $S_{11}=264$ , so the least count is 11.

Turn value bounds into term positions

“Terms between 20 and 50” refers to their values, not their positions. Find the first and last allowed terms before adding them.

Worked example

Add the terms of $5,11,17,\ldots$ strictly between 20 and 50.

Here $u_n=6n-1$ . “Strictly between” excludes both boundaries:

\begin{gathered}20<6n-1<50\\3.5<n<8.5\end{gathered}

The integer positions are 4 through 8, giving first term 23 and last term 47. Count both ends: $8-4+1=5$ terms.

S=\frac52(23+47)=175

For inclusive bounds, use $\le$ . If the difference is negative, dividing an inequality by it reverses the sign. A short calculator table of $u_n$ can check the two endpoint positions.

When the sum gives the number of terms

Worked example

How many terms of $2,5,8,\ldots$ sum to 155?

Use the sum formula because 155 is a total:

\frac n2[4+3(n-1)]=155

3n^2+n-310=0

The polynomial solver gives 10 and $-31/3$ . These convenient roots give the working $(n-10)(3n+31)=0$ . Only $n=10$ is a positive integer.

Common mistake

Do not round a non-integer solution of an exact sum equation to make it a term count. It may mean no such count exists.

Now you try

Your turnIndependent practice

(a) Add all multiples of 7 strictly between 50 and 100.

(b) Find the least number of terms of $4,8,12,\ldots$ whose sum exceeds 220.

Show worked answer

(a) The terms run from 56 to 98, seven terms:

S=\frac72(56+98)=539

(b) $S_n=2n(n+1)$ . The sums increase since every term is positive. $S_{10}=220$ and $S_{11}=264$ , so the least count is 11.