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Arithmetic Sequence
The Core Identity
An Arithmetic Sequence is a list of numbers where each term is found by adding a constant common difference to the previous one, creating linear growth.
$$d = a_{n} - a_{n-1}$$
The common difference is the gap between any two consecutive terms.
$d$=Common difference(unitless)
$a_{n}$=Any term in the sequence(unitless)
$a_{n-1}$=The term immediately preceding $a_{n}$(unitless)
$d > 0$
→The sequence is increasing.
$d < 0$
→The sequence is decreasing.
$d = 0$
→All terms are equal (constant sequence).
Linear Growth: Because the change is constant, the terms follow a straight-line pattern when plotted against index $n$.
Consecutive Check: The difference must be identical for ALL pairs of neighbors — check at least two pairs to confirm an A.P.
Non-Consecutive Shortcut: For any two terms $a_m$ and $a_p$, the common difference is $d = \frac{a_p - a_m}{p - m}$.
The General Term
The general term allows you to calculate the value of any specific index without listing every intermediate number.
$$a_n = a_1 + (n - 1)d$$
The n-th term is the starting value plus $(n-1)$ jumps of size $d$.
$a_n$=Value of the n-th term(unitless)
$a_1$=First term (starting value)(unitless)
$n$=Position (index) of the term(positive integer)
$d$=Common difference(unitless)
$n = 1$
→$a_1 = a_1 + 0 \cdot d = a_1$ (zero jumps from the start).
Linear Function Form: $a_n = dn + (a_1 - d)$ — this is a linear function of $n$ with slope $d$ and y-intercept $(a_1 - d)$.
Finding $n$ for a Given Term: Rearrange to $n = \frac{a_n - a_1}{d} + 1$. If this isn't a positive integer, the value is not in the sequence.
Shifted Index Problems: If given $a_{n-k} = f(n)$, substitute values of $n$ that make the subscript equal to 1, 2, 3... to find the actual first terms.
Arithmetic Series (Sum of Terms)
An Arithmetic series is the partial sum of the terms in a sequence. When you know the first and last terms, use Gauss's pairing method.
$$S_n = \frac{n}{2}(a_1 + a_n)$$
The sum equals the average of the first and last terms, multiplied by the number of terms.
$S_n$=Sum of the first $n$ terms(unitless)
$n$=Number of terms to sum(positive integer)
$a_1$=First term(unitless)
$a_n$=Last term in the sum(unitless)
Gauss's Insight: Pair the first term with the last, the second with the second-last, etc. Each pair sums to $a_1 + a_n$, and there are $\frac{n}{2}$ such pairs.
When to Use: Choose this form when you already know (or can easily find) $a_n$.
When the last term $a_n$ is unknown, substitute $a_n = a_1 + (n-1)d$ into the quick formula to get the standard form.
$$S_n = \frac{n}{2}[2a_1 + (n-1)d]$$
Expresses the sum entirely in terms of $a_1$, $d$, and $n$ — no need to find $a_n$ first.
$S_n$=Sum of the first $n$ terms(unitless)
$a_1$=First term(unitless)
$d$=Common difference(unitless)
$n$=Number of terms(positive integer)
When to Use: Choose this form when you know $a_1$ and $d$ but not $a_n$.
Quadratic in $n$: $S_n = \frac{d}{2}n^2 + (a_1 - \frac{d}{2})n$ — the sum of an A.P. is always a quadratic function of $n$.
The Arithmetic Mean
The arithmetic mean is the midpoint between two numbers; when inserted between them, it creates a three-term arithmetic sequence.
$$A = \frac{a + b}{2}$$
The A.M. of two numbers is their average — equidistant from both.
$A$=Arithmetic mean(unitless)
$a, b$=The two endpoint numbers(unitless)
Equidistant Property: $A - a = b - A = d$, so the common difference is $d = \frac{b - a}{2}$.
General A.M. in a Sequence: Any term $a_n$ is the arithmetic mean of its two neighbours: $a_n = \frac{a_{n-1} + a_{n+1}}{2}$.
To insert $n$ arithmetic means between two numbers $a$ and $b$, construct an A.P. of $(n+2)$ terms with $a$ as the first term and $b$ as the last.
$$d = \frac{b - a}{n + 1}$$
The $(n+2)$-term A.P. has $(n+1)$ gaps of equal size between $a$ and $b$.
$d$=Common difference of the extended A.P.(unitless)
$a$=First endpoint (= $a_1$)(unitless)
$b$=Last endpoint (= $a_{n+2}$)(unitless)
$n$=Number of means to insert(positive integer)
Denominator is $n+1$: With $n$ means, there are $n + 2$ total terms and $n + 1$ gaps.
The $k$-th Mean: $A_k = a + k \cdot \frac{b-a}{n+1}$ for $k = 1, 2, \ldots, n$.
Sum of $n$ AMs: The sum of all $n$ arithmetic means between $a$ and $b$ equals $n \times \frac{a+b}{2}$ (i.e., $n$ times their A.M.).
Application Strategy
Worked Examples
Problem Statement
Find the 50th term of the arithmetic sequence: 5, 12, 19, 26...
General Term FormulaCommon Difference Identification
Strategic Toolkit
$$a_n = a_1 + (n - 1)d$$
Step 1
Identify Knowns
The sequence starts with 5, so $a_1 = 5$. The common difference is $12 - 5 = 7$, so $d = 7$.
$a_1 = 5, \; d = 7, \; n = 50$
Step 2
Substitute
Plug into the general term formula.
$a_{50} = 5 + (50 - 1) \times 7 = 5 + 49 \times 7$
Step 3
Calculate
Simplify: multiply before adding the first term.
$a_{50} = 5 + 343 = 348$
Problem Statement
Find the sum of all multiples of 3 between 1 and 100.
Sequence IdentificationFinding Number of TermsSum of Arithmetic Series
Strategic Toolkit
$$S_n = \frac{n}{2}(a_1 + a_n)$$
Step 1
Define the Sequence
Multiples of 3 between 1 and 100: $3, 6, 9, \ldots, 99$. So $a_1 = 3$, $a_n = 99$, $d = 3$.
$a_1 = 3, \; a_n = 99, \; d = 3$
Step 2
Find $n$
Use the general term formula to count terms.
$99 = 3 + (n - 1)(3) \implies 96 = 3(n-1) \implies n = 33$
Step 3
Calculate the Sum
Apply the quick sum formula with first and last terms known.
$S_{33} = \frac{33}{2}(3 + 99) = \frac{33}{2}(102) = 33 \times 51 = 1683$
Problem Statement
If the 5th term of an A.P. is 13 and the 17th term is 49, find $a_n$ and $a_{13}$.
Simultaneous EquationsGeneral Term Construction
Strategic Toolkit
$$a_n = a_1 + (n-1)d$$
Step 1
Set Up Equations
$a_5 = a_1 + 4d = 13$ ... (i) and $a_{17} = a_1 + 16d = 49$ ... (ii).
$a_1 + 4d = 13, \quad a_1 + 16d = 49$
Step 2
Subtract to Eliminate $a_1$
(ii) $-$ (i): $12d = 36$, so $d = 3$.
$12d = 36 \implies d = 3$
Step 3
Back-Substitute
From (i): $a_1 = 13 - 4(3) = 1$.
$a_1 = 1$
Step 4
Write the Results
The general term is $a_n = 1 + (n-1)(3) = 3n - 2$. So $a_{13} = 3(13) - 2 = 37$.
$a_n = 3n - 2, \quad a_{13} = 37$
Problem Statement
Find three A.Ms between $\sqrt{2}$ and $3\sqrt{2}$.
n Arithmetic MeansCommon Difference from Endpoints
Strategic Toolkit
$$d = \frac{b - a}{n + 1}$$
Step 1
Identify Parameters
$a = \sqrt{2}$, $b = 3\sqrt{2}$, $n = 3$ means. Total terms: $3 + 2 = 5$.
$a_1 = \sqrt{2}, \; a_5 = 3\sqrt{2}$
Step 2
Find $d$
$d = \frac{3\sqrt{2} - \sqrt{2}}{3 + 1} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
$d = \frac{1}{\sqrt{2}}$
Step 3
List the Means
$A_1 = \sqrt{2} + \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$, $A_2 = \frac{4}{\sqrt{2}} = 2\sqrt{2}$, $A_3 = \frac{5}{\sqrt{2}}$.
$\frac{3}{\sqrt{2}}, \; 2\sqrt{2}, \; \frac{5}{\sqrt{2}}$
Problem Statement
A man repays a loan of Rs. 1120 by paying Rs. 15 in the first month and increasing by Rs. 10 each month. How many months to clear the loan?
Sum FormulaQuadratic Equation from Series
Strategic Toolkit
$$S_n = \frac{n}{2}[2a_1 + (n-1)d]$$
Step 1
Set Up
$a_1 = 15$, $d = 10$, $S_n = 1120$.
$1120 = \frac{n}{2}[2(15) + (n-1)(10)]$
Step 2
Simplify
$1120 = \frac{n}{2}[30 + 10n - 10] = 5n(n + 2)$.
$n(n+2) = 224 \implies n^2 + 2n - 224 = 0$
Step 3
Solve
Discriminant: $\sqrt{4 + 896} = \sqrt{900} = 30$. So $n = \frac{-2 + 30}{2} = 14$. (Reject $n = -16$.)
$n = 14$ months
Memorize formulas
General Term of A.P.
Find any term from its position, the first term, and the common difference.
Formula
$a_n = a_1 + (n - 1)d$
Sum (Standard Form)
Sum when $a_n$ is unknown — uses $a_1$, $d$, $n$ only.
Formula
$S_n = \frac{n}{2}[2a_1 + (n - 1)d]$
Sum (Gauss Form)
Sum when first and last terms are known.
Formula
$S_n = \frac{n}{2}(a_1 + a_n)$
Arithmetic Mean
The midpoint between two values $a$ and $b$.
Formula
$A = \frac{a + b}{2}$
Common Difference for $n$ AMs
When inserting $n$ arithmetic means between $a$ and $b$.
Formula
$d = \frac{b - a}{n + 1}$
Review
1
In an A.P., the sum of the 3rd and 8th terms is 29, and the sum of the 6th and 11th terms is 47. Find the first term.
Ans
1
Review
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Write the two term-sum equations
In an A.P., $a_n = a_1 + (n-1)d$. Use that to express the two given sums in terms of $a_1$ and $d$.
$$2a_1 + 9d = 29, \qquad 2a_1 + 15d = 47$$
Eliminate the first term
Subtract the first equation from the second to isolate the common difference, then substitute back to get the first term.
$$6d = 18 \implies d = 3$$
$$2a_1 + 9(3) = 29 \implies a_1 = 1$$
2
A sequence is defined as 100, 93, 86, 79... Which term is the first negative term?
Ans
The 16th term
Review
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Form the nth-term inequality
The sequence is an A.P. with first term 100 and common difference $-7$, so the nth term is $a_n = 100 - 7(n-1)$.
$$a_n = 100 - 7(n-1)$$
Find the first negative term
Solve $a_n < 0$ and take the first whole-number index that satisfies the inequality.
$$100 - 7(n-1) < 0 \implies n > 15.28 \implies n = 16$$
3
The angles of a triangle are in A.P. If the smallest angle is half the largest angle, find the three angles.
Ans
40°, 60°, 80°
Review
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Represent the triangle angles in A.P.
Write the three angles as $a-d$, $a$, and $a+d$ so the middle angle is the average of the other two.
$$a-d,\ a,\ a+d$$
Use the angle sum and the ratio condition
The triangle sum gives $3a = 180°$, and the smallest being half the largest gives a second equation in $a$ and $d$.
$$a = 60°, \quad a-d = \frac{a+d}{2} \implies d = 20°$$
Substitute the values
Insert $a = 60°$ and $d = 20°$ into the A.P. form to get the three angles.
$$40°,\ 60°,\ 80°$$
4
Find the sum of all multiples of 7 from 56 to 350 inclusive.
Ans
8729
Review
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Count the terms in the sequence
The multiples of 7 form an A.P. with first term 56, last term 350, and common difference 7.
$$n = \frac{350 - 56}{7} + 1 = 43$$
Apply the A.P. sum formula
Use the sum formula for an arithmetic sequence with the first and last terms already known.
$$S_{43} = \frac{43}{2}(56 + 350) = 8729$$
5
If $3a_7 = 7a_4$ and $a_{10} = 33$ in an A.P., find $a_1$.
Ans
$-3$
Review
Tap to reveal
Express the given terms in A.P. form
Write each required term using $a_n = a_1 + (n-1)d$ so the condition can be turned into an equation in $a_1$ and $d$.
$$a_7 = a_1 + 6d, \quad a_4 = a_1 + 3d$$
Use the ratio condition to link $a_1$ and $d$
Substitute the term expressions into $3a_7 = 7a_4$ and simplify.
$$3(a_1 + 6d) = 7(a_1 + 3d) \implies 4a_1 = -3d$$
Substitute into the 10th-term condition
Use $a_{10} = a_1 + 9d = 33$ together with the relation above to solve for the first term.
$$a_1 + 9\left(-\frac{4a_1}{3}\right) = 33 \implies a_1 = -3$$
6
Three A.Ms are inserted between 3 and 11. What is the second mean?
Ans
7
Review
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Find the common difference
Three means inserted between 3 and 11 create a 5-term A.P., so the gap is split into four equal parts.
$$d = \frac{11 - 3}{4} = 2$$
Read off the second inserted mean
Starting from 3 and adding the common difference twice gives the second mean.
$$3 + 2d = 7$$
7
A contractor stacks pipes in rows. The bottom row has 30 pipes, the next has 28, then 26, and so on, decreasing by 2 each row. If the top row has 8 pipes, how many pipes are in the stack?
Ans
228
Review
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Count the number of rows
The rows form an A.P. from 30 down to 8 with common difference $-2$, so determine how many terms are present.
$$n = \frac{8 - 30}{-2} + 1 = 12$$
Use the sum formula
Add the first and last row counts and multiply by half the number of rows.
$$S_{12} = \frac{12}{2}(30 + 8) = 228$$
8
If $a$, $b$, $c$ are in A.P. (all non-zero), then $\frac{1}{bc}$, $\frac{1}{ca}$, $\frac{1}{ab}$ are in:
Ans
A.P.
Review
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Use the A.P. condition on $a$, $b$, and $c$
For numbers in A.P., the middle term is the average of the other two, so $2b = a + c$.
$$2b = a + c$$
Divide through by $abc$
Rewrite the relation in reciprocal form to match the three given terms.
$$\frac{2}{ac} = \frac{1}{bc} + \frac{1}{ab}$$
Recognize the new A.P.
The middle reciprocal term is the average of the first and third reciprocal terms, so the three reciprocals are in A.P.
$$\frac{1}{bc},\ \frac{1}{ca},\ \frac{1}{ab} \text{ are in A.P.}$$
9
Determine whether $-19$ is a term of the A.P. $17, 13, 9, \ldots$
Ans
Yes, it is the 10th term
Review
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Write the nth-term formula
The sequence is an A.P. with first term 17 and common difference $-4$.
$$a_n = 17 - 4(n-1)$$
Check whether -19 occurs for a positive index
Set the nth term equal to -19 and solve for $n$.
$$17 - 4(n-1) = -19 \implies n = 10$$
Interpret the index
Because $n = 10$ is a positive integer, -19 is a term of the sequence.
$$-19 \text{ is the 10th term}$$
10
If $l$, $m$, $n$ are the $p$th, $q$th, and $r$th terms of an A.P. respectively, what is the value of $l(q - r) + m(r - p) + n(p - q)$?
Ans
0
Review
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Write each term of the A.P.
Let the sequence start at $a$ with common difference $d$, so each indexed term can be written directly in terms of its position.
$$l = a + (p-1)d,\ m = a + (q-1)d,\ n = a + (r-1)d$$
Substitute and group like terms
After substitution, the coefficients of $a$ cancel because they sum to zero, and the coefficients of $d$ also cancel after expansion.
$$l(q-r) + m(r-p) + n(p-q) = 0$$
11
An investor earned Rs. 6000 in year 1 and Rs. 12000 in year 11 on the same investment. If earnings increase uniformly each year, what is the total income over 11 years?
Ans
Rs. 99000
Review
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Identify the arithmetic series endpoints
The yearly incomes form an A.P. with first term 6000, last term 12000, and 11 terms total.
$$a_1 = 6000, \quad a_{11} = 12000, \quad n = 11$$
Apply the sum formula
Use the arithmetic-series sum formula with the first and last terms.
$$S_{11} = \frac{11}{2}(6000 + 12000) = 99000$$
12
Find the $n$-th term of the sequence: $\left(\frac{4}{3}\right)^2, \left(\frac{7}{3}\right)^2, \left(\frac{10}{3}\right)^2, \ldots$
Ans
$\left(\frac{3n+1}{3}\right)^2$
Review
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Find the pattern in the base terms
Ignore the squares for a moment: the numbers inside them form an A.P. with first term $\frac{4}{3}$ and common difference 1.
$$\frac{4}{3},\ \frac{7}{3},\ \frac{10}{3},\ldots$$
Write the nth base term
Use the A.P. formula to get the nth value inside the square, then square it to match the sequence.
$$\text{nth base} = \frac{4}{3} + (n-1) = \frac{3n+1}{3}$$
$$a_n = \left(\frac{3n+1}{3}\right)^2$$
Question 1 of 12
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1
In an A.P., the sum of the 3rd and 8th terms is 29, and the sum of the 6th and 11th terms is 47. Find the first term.