CSCA MATHEMATICS · MODULE 02 · LESSON 02.10

General Terms & Finite Sums

Understand individual sequence terms, sigma notation, arithmetic and geometric partial sums, and applications through step-by-step worked examples.

Arithmetic sumsGeometric sumsSigma notation
Lesson format

This is the written teaching lesson. Standalone practice questions and their answer explanations have been omitted for a future Tutor LMS quiz. Worked teaching examples remain included.

1. General terms and sums: What is the difference?

A general term describes an individual term of a sequence. For example:

\[a_n=3n+1\]

This formula gives:

\[a_1=4, \qquad a_2=7, \qquad a_3=10\]

A sum adds several terms together. We use \(S_n\) to represent the sum of the first \(n\) terms:

\[S_n=a_1+a_2+\cdots+a_n\]

For the sequence above:

\[S_3=4+7+10=21\]

Notice the difference:

\[a_3=10,\]

but:

\[S_3=21\]

The third term and the sum of the first three terms are different quantities.

Terms and partial sums

Notation Meaning Example for \(4,7,10,\ldots\)
\(a_n\) The term in position \(n\) \(a_3=10\)
\(S_n\) The sum of the first \(n\) terms \(S_3=21\)
\(n\) A term index or number of terms, according to context \(n=3\)

Recovering a term from sums

For \(n\geq2\) :

\[a_n=S_n-S_{n-1}\]

Why?

\[S_n=a_1+\cdots+a_{n-1}+a_n,\]

whereas:

\[S_{n-1}=a_1+\cdots+a_{n-1}\]

Subtracting leaves only \(a_n\) . For the first term:

\[a_1=S_1\]

Worked example 1

Suppose:

\[S_8=100, \qquad S_7=79\]

Then:

\[a_8=S_8-S_7=100-79=21\]

This relationship applies to any numerical sequence, not only arithmetic or geometric sequences.

2. Reviewing general term formulas

Arithmetic sequences

An arithmetic sequence has a constant difference:

\[d=a_{n+1}-a_n\]

Its general term is:

\[a_n=a_1+(n-1)d\]

Example:

\[5,\ 8,\ 11,\ 14,\ldots\]

has:

\[a_1=5, \qquad d=3\]

Therefore:

\[a_n=5+3(n-1)=3n+2\]

Geometric sequences

A geometric sequence is generated by multiplying each term by a fixed common ratio \(r\) :

\[a_{n+1}=ra_n\]

Its general term is:

\[a_n=a_1r^{n-1}\]

When the preceding term is nonzero, the ratio can be calculated as:

\[r=\frac{a_{n+1}}{a_n}\]

Example:

\[3,\ 6,\ 12,\ 24,\ldots\]

has:

\[a_1=3, \qquad r=2\]

Therefore:

\[a_n=3\cdot2^{n-1}\]

Key distinction: Arithmetic sequences use repeated addition. Geometric sequences use repeated multiplication.

3. Sigma notation

The symbol:

\[\sum\]

is the Greek capital letter sigma. It means “add the indicated terms.” For example:

\[\sum_{k=1}^{4}a_k =a_1+a_2+a_3+a_4\]

In this expression: \(k\) is the summation index. \(1\) is the starting index. \(4\) is the ending index. \(a_k\) is the expression to evaluate and add.

Worked example 2: Expanding a sum

Evaluate:

\[\sum_{k=1}^{4}(2k+1)\]

Substitute:

\[k=1,2,3,4\]

Then:

\[\sum_{k=1}^{4}(2k+1) =(2\cdot1+1)+(2\cdot2+1)+(2\cdot3+1)+(2\cdot4+1)\]

Therefore:

\[\sum_{k=1}^{4}(2k+1)=3+5+7+9=24\]

Sums do not always start at one

For example:

\[\sum_{k=3}^{6}k =3+4+5+6 =18\]

The number of terms from index \(p\) through index \(q\) including both endpoints, is:

\[q-p+1\]

Here:

\[6-3+1=4\]

The index letter can change

These expressions mean the same thing:

\[\sum_{k=1}^{n}a_k\]

and:

\[\sum_{j=1}^{n}a_j\]

The index letter is a placeholder.

Useful summation rules

A constant added \(n\) times gives:

\[\sum_{k=1}^{n}c=nc\]

A constant factor can be taken outside a sum:

\[\sum_{k=1}^{n}cb_k =c\sum_{k=1}^{n}b_k\]

A sum of expressions can be separated:

\[\sum_{k=1}^{n}(b_k+c_k) = \sum_{k=1}^{n}b_k+\sum_{k=1}^{n}c_k\]

These rules also apply to differences.

4. Sum of an arithmetic sequence

For an arithmetic sequence:

\[S_n=\frac{n(a_1+a_n)}{2}\]

In words: Multiply the number of terms by the average of the first and last terms.

Why does this formula work?

Write the sum in forward order:

\[S_n=a_1+a_2+\cdots+a_n\]

Write it again in reverse order:

\[S_n=a_n+a_{n-1}+\cdots+a_1\]

In an arithmetic sequence, corresponding pairs have the same sum:

\[a_1+a_n=a_2+a_{n-1}\]

Adding the two lines gives \(n\) copies of that pair sum:

\[2S_n=n(a_1+a_n)\]

Therefore:

\[S_n=\frac{n(a_1+a_n)}{2}\]

This argument works whether \(n\) is even or odd.

An equivalent formula

Since:

\[a_n=a_1+(n-1)d,\]

substitute into the sum formula:

\[S_n=\frac{n}{2}\bigl(a_1+a_1+(n-1)d\bigr)\]

Therefore:

\[S_n=\frac{n}{2}\bigl(2a_1+(n-1)d\bigr)\]

Another equivalent form is:

\[S_n=na_1+\frac{n(n-1)}{2}d\]

Choose the form that matches the information provided.

Worked example 3: First and last terms known

Find the sum of:

\[7,\ 11,\ 15,\ldots,\ 63\]

Step 1: Find the number of terms.

\[63=7+(n-1)4\]
\[56=4(n-1)\]
\[n=15\]

Step 2: Find the sum.

\[S_{15}=\frac{15(7+63)}{2}\]
\[S_{15}=\frac{15\cdot70}{2}=525\]

Worked example 4: First term and difference known

Find the sum of the first \(20\) terms when:

\[a_1=5, \qquad d=3\]

Use:

\[S_n=\frac{n}{2}\bigl(2a_1+(n-1)d\bigr)\]

Then:

\[S_{20}=\frac{20}{2}\bigl(2(5)+19(3)\bigr)\]
\[S_{20}=10(10+57)=670\]

A useful special case

For:

\[1+2+3+\cdots+n,\]

the first term is \(1\) the last term is \(n\) and there are \(n\) terms. Therefore:

\[\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\]

For example:

\[1+2+\cdots+100 =\frac{100(101)}{2} =5050\]

5. Sum of a geometric sequence

For a geometric sequence:

\[a_1,\ a_1r,\ a_1r^2,\ldots,\ a_1r^{n-1},\]

the finite sum is:

\[S_n=\frac{a_1(1-r^n)}{1-r}, \qquad r\neq1\]

An equivalent form is:

\[S_n=\frac{a_1(r^n-1)}{r-1}, \qquad r\neq1\]

Both formulas give the same result because multiplying the numerator and denominator by \(-1\) does not change a fraction.

Why does the formula work?

Write:

\[S_n=a_1+a_1r+\cdots+a_1r^{n-1}\]

Multiply by \(r\) :

\[rS_n=a_1r+a_1r^2+\cdots+a_1r^n\]

Subtract the second equation from the first. The middle terms cancel:

\[S_n-rS_n=a_1-a_1r^n\]

Factor:

\[(1-r)S_n=a_1(1-r^n)\]

For \(r\neq1\) divide:

\[S_n=\frac{a_1(1-r^n)}{1-r}\]

Special case: \(r=1\)

Every term equals \(a_1\) so:

\[S_n=na_1\]

Do not use a formula with denominator \(1-r\) when \(r=1\)

Worked example 5: Ratio greater than one

Find the sum of the first six terms of:

\[3,\ 6,\ 12,\ldots\]

Here:

\[a_1=3, \qquad r=2, \qquad n=6\]

Use:

\[S_6=\frac{3(2^6-1)}{2-1}\]
\[S_6=3(64-1)=189\]

Check:

\[3+6+12+24+48+96=189\]

Worked example 6: A fractional ratio

Find the sum of the first four terms of:

\[16,\ 8,\ 4,\ 2,\ldots\]

Here:

\[a_1=16, \qquad r=\frac12\]

Then:

\[S_4= \frac{16\left(1-\left(\frac12\right)^4\right)} {1-\frac12}\]
\[S_4= \frac{16\left(1-\frac1{16}\right)} {\frac12}\]
\[S_4=\frac{15}{\frac12}=30\]

Worked example 7: A negative ratio

Find the sum of the first five terms of:

\[2,\ -6,\ 18,\ -54,\ 162,\ldots\]

Here:

\[a_1=2, \qquad r=-3\]

Therefore:

\[S_5=\frac{2(1-(-3)^5)}{1-(-3)}\]

Since:

\[(-3)^5=-243,\]

we obtain:

\[S_5=\frac{2(244)}{4}=122\]

The finite geometric sum formula works for negative ratios too. Keep negative bases inside parentheses.

6. Choosing the right formula

General terms and finite sums

Sequence type General term Sum of first \(n\) terms
Arithmetic \(a_n=a_1+(n-1)d\) \(S_n=\frac{n(a_1+a_n)}{2}\)
Arithmetic: equivalent form \(a_n=a_1+(n-1)d\) \(S_n=\frac{n}{2}(2a_1+(n-1)d)\)
Geometric, \(r\neq1\) \(a_n=a_1r^{n-1}\) \(S_n=\frac{a_1(1-r^n)}{1-r}\)
Geometric: equivalent form, \(r\neq1\) \(a_n=a_1r^{n-1}\) \(S_n=\frac{a_1(r^n-1)}{r-1}\)
Constant sequence \(a_n=a_1\) \(S_n=na_1\)

Summing a section of a sequence

To add terms from \(a_p\) through \(a_q\) where \(p\geq2\)

\[a_p+a_{p+1}+\cdots+a_q=S_q-S_{p-1}\]

For example:

\[a_5+a_6+\cdots+a_{12}=S_{12}-S_4\]

There are:

\[12-5+1=8\]

terms in this section.

7. Finding the number of terms from a sum

The number of terms must be a positive integer.

Worked example 8: Arithmetic sum

How many terms of:

\[3,\ 5,\ 7,\ldots\]

have a sum of \(120\) ? Here:

\[a_1=3, \qquad d=2\]

Use:

\[120=\frac{n}{2}\bigl(2(3)+(n-1)2\bigr)\]

Simplify:

\[120=\frac{n}{2}(2n+4)\]
\[120=n(n+2)\]

Therefore:

\[n^2+2n-120=0\]

Factor:

\[(n+12)(n-10)=0\]

The candidates are:

\[n=-12 \quad\text{or}\quad n=10\]

Only the positive integer is valid:

\[n=10\]

Worked example 9: Geometric sum

How many terms of:

\[2,\ 6,\ 18,\ldots\]

have a sum of \(242\) ? Here:

\[a_1=2, \qquad r=3\]

Then:

\[242=\frac{2(3^n-1)}{3-1}\]

Simplify:

\[242=3^n-1\]
\[3^n=243\]

Since:

\[243=3^5,\]

we obtain:

\[n=5\]

Can a sum give more than one valid answer?

Yes. If terms change sign, partial sums need not keep increasing. For example, the arithmetic sequence:

\[3,\ 1,\ -1,\ -3,\ldots\]

has:

\[S_1=3\]

and:

\[S_3=3+1-1=3\]

Therefore, check every positive integer candidate rather than assuming the answer is unique.

8. Finding unknown sequence parameters

Worked example 10: Unknown first term

An arithmetic sequence has:

\[d=3, \qquad S_{10}=185\]

Find \(a_1\) .

\[185=\frac{10}{2}\bigl(2a_1+9(3)\bigr)\]
\[185=5(2a_1+27)\]

Divide by \(5\) :

\[37=2a_1+27\]

Therefore:

\[a_1=5\]

Worked example 11: Unknown common difference

An arithmetic sequence has:

\[a_1=4, \qquad S_8=116\]

Find \(d\)

\[116=\frac82\bigl(2(4)+7d\bigr)\]
\[116=4(8+7d)\]
\[29=8+7d\]

Therefore:

\[d=3\]

Worked example 12: Unknown first term of a geometric sequence

A geometric sequence has:

\[r=2, \qquad S_5=93\]

Find \(a_1\)

\[93=\frac{a_1(2^5-1)}{2-1}\]
\[93=31a_1\]

Therefore:

\[a_1=3\]

Worked example 13: Unknown common ratio

A geometric sequence has:

\[a_1=2, \qquad S_3=14\]

Find all real possible ratios. Write the three terms directly:

\[2,\ 2r,\ 2r^2\]

Their sum gives:

\[2+2r+2r^2=14\]

Divide by \(2\)

\[1+r+r^2=7\]
\[r^2+r-6=0\]

Factor:

\[(r+3)(r-2)=0\]

Therefore:

\[r=-3 \quad\text{or}\quad r=2\]

Check both:

\[2-6+18=14,\]
\[2+4+8=14\]

Both ratios are valid. A negative ratio must not be rejected unless the question imposes an additional condition, such as all terms being positive.

9. Mixed arithmetic and geometric questions

Begin by identifying the sequence type.

Worked example 14: Compare two sums

Find and compare the sums of the first five terms of: Sequence A:

\[2,\ 5,\ 8,\ 11,\ldots\]

Sequence B:

\[2,\ 4,\ 8,\ 16,\ldots\]

For Sequence A:

\[a_1=2, \qquad d=3\]
\[S_5=\frac52\bigl(2(2)+4(3)\bigr)\]
\[S_5=\frac52(16)=40\]

For Sequence B:

\[a_1=2, \qquad r=2\]
\[S_5=\frac{2(2^5-1)}{2-1}=62\]

Therefore, Sequence B has the larger five-term sum by:

\[62-40=22\]

Worked example 15: Using sigma notation

Evaluate:

\[\sum_{k=1}^{5}\left(3k+2^{k-1}\right)\]

Separate the sum:

\[\sum_{k=1}^{5}3k+\sum_{k=1}^{5}2^{k-1}\]

The first part is:

\[3(1+2+3+4+5) =3\cdot\frac{5(6)}2 =45\]

The second part is geometric:

\[1+2+4+8+16 =\frac{2^5-1}{2-1} =31\]

Therefore:

\[\sum_{k=1}^{5}\left(3k+2^{k-1}\right)=45+31=76\]

The combined sequence need not itself be arithmetic or geometric. We can still sum its arithmetic and geometric components separately.

10. Practical finite-summation problems

Worked example 16: Total seats

An auditorium has \(12\) rows. The first row has \(18\) seats, and each following row has \(3\) more seats. Find the total number of seats. First find the final row:

\[a_{12}=18+11(3)=51\]

Then:

\[S_{12}=\frac{12(18+51)}2\]
\[S_{12}=6(69)=414\]

The auditorium has:

\[414\text{ seats}\]

Worked example 17: Increasing weekly savings

A student saves \(50\) RMB in week \(1\) then saves \(10\) RMB more each following week. Find the total saved during eight weeks.

\[a_1=50, \qquad d=10, \qquad n=8\]
\[S_8=\frac82\bigl(2(50)+7(10)\bigr)\]
\[S_8=4(170)=680\]

The total saved is:

\[680\text{ RMB}\]

The amount saved in week \(8\) alone is \(120\) RMB. Do not confuse this with the total.

Worked example 18: Finite production growth

A workshop produces \(100\) items in month \(1\) Production increases by \(20\%\) each month under the model. Find total production over the first three months. The monthly amounts form a geometric sequence with:

\[a_1=100, \qquad r=1.20\]

The three amounts are:

\[100,\ 120,\ 144\]

Therefore:

\[S_3=100+120+144=364\]

Using the formula gives the same result:

\[S_3=\frac{100(1.2^3-1)}{1.2-1}=364\]

The model predicts a total of:

\[364\text{ items}\]

Modeling reminder: Equal added amounts suggest an arithmetic sequence. Equal percentage changes suggest a geometric sequence.

11. Common mistakes

Mistake 1: Using a term formula to answer a sum question

\[a_n=a_1+(n-1)d\]

finds one arithmetic term. To add the first \(n\) terms, use an appropriate sum formula.

Mistake 2: Confusing geometric exponents

The final term is:

\[a_n=a_1r^{n-1}\]

The geometric sum formula contains:

\[r^n\]

These exponents serve different purposes.

Mistake 3: Counting sigma terms incorrectly

The sum:

\[\sum_{k=3}^{8}a_k\]

contains:

\[8-3+1=6\]

terms, not five or eight.

Mistake 4: Forgetting the case \(r=1\)

When:

\[r=1,\]

use:

\[S_n=na_1\]

Mistake 5: Rejecting negative ratios automatically

A geometric sequence may alternate in sign. Check the problem’s conditions before rejecting a negative ratio.

Mistake 6: Rounding a calculated index

If an exact sum requires:

\[n=6.4,\]

there is no valid integer number of terms giving that exact sum. Do not round unless the question asks for a threshold, approximation, or minimum number.

Mistake 7: Applying an infinite-sum formula to a finite question

A finite sum and an infinite sum are different quantities. Use the specified number of terms.

12. Optional enrichment: Infinite geometric series

This section is separate from the finite-summation skills required in the main lesson. For a geometric sequence with:

\[|r|<1,\]

the finite sums approach a limiting value as more terms are added. Because:

\[r^n\rightarrow0,\]

the finite formula:

\[S_n=\frac{a_1(1-r^n)}{1-r}\]

approaches:

\[S_\infty=\frac{a_1}{1-r}\]

Example

Consider:

\[8+4+2+1+\cdots\]

Here:

\[a_1=8, \qquad r=\frac12\]

Therefore:

\[S_\infty=\frac8{1-\frac12}=16\]

The sum of the first four terms is only:

\[S_4=8+4+2+1=15\]

The finite sum and infinite limiting sum are different. For a nonzero first term, the infinite geometric series has a finite sum only when:

\[|r|<1\]

13. Lesson review

You should now be able to: Distinguish an individual term \(a_n\) from a partial sum \(S_n\) Expand and evaluate finite sums written in sigma notation. Choose an arithmetic or geometric sum formula. Count terms correctly when a sum starts at an index other than \(1\) Determine unknown first terms, differences, ratios, and numbers of terms. Check all candidate answers against the sequence’s conditions. Interpret practical questions as individual-term or total-sum problems. The main finite-sum formulas are:

\[S_n=\frac{n(a_1+a_n)}2\]

for arithmetic sequences, and:

\[S_n=\frac{a_1(1-r^n)}{1-r}, \qquad r\neq1,\]

for geometric sequences. Transition to Phase D: Sequence sums add values at discrete positions. The next phase introduces calculus by examining how functions change, beginning with average rates of change and the idea of a derivative.

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