This sheet collects the Sequences and Series Class 11 formulas you will need for quick revision: sequences, series, the general term and sum of a geometric progression (G.P.), geometric mean, and the A.M.–G.M. relationship. Every formula follows the Rationalised NCERT Class 11 Mathematics textbook — verify any line in the chapter PDF on ncert.nic.in.
Each formula is grouped by topic, with a symbol-meaning table, when-to-use guidance, and three worked examples using fresh numbers. For the explanations and derivations this sheet deliberately skips, see the Class 11 Maths Formulas section.
Formulas at a Glance
The table below is the whole chapter in one screen — every formula on this page, in the order you will reach for it.
| Purpose (what you are finding) | Formula |
|---|---|
| Write a series in compact form (sigma notation) | \( \sum_{k=1}^{n} a_k = a_1 + a_2 + \dots + a_n \) |
| Test whether a sequence is a G.P. | \( \frac{a_{k+1}}{a_k} = r,\ k \geq 1 \) |
| General (nth) term of a G.P. | \( a_n = ar^{n-1} \) |
| Sum of first n terms of a G.P., when \( r = 1 \) | \( S_n = na \) |
| Sum of first n terms of a G.P., when \( r \neq 1 \) | \( S_n = \frac{a(r^n-1)}{r-1} = \frac{a(1-r^n)}{1-r} \) |
| Geometric mean of two positive numbers | \( G = \sqrt{ab} \) |
| Common ratio when n numbers are inserted between a and b | \( b = ar^{n+1} \Rightarrow r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}} \) |
| k-th inserted geometric mean | \( G_k = a\left(\frac{b}{a}\right)^{\frac{k}{n+1}} \) |
| Arithmetic mean of two numbers | \( A = \frac{a+b}{2} \) |
| A.M.–G.M. relationship for positive numbers | \( A – G = \frac{(\sqrt{a}-\sqrt{b})^2}{2} \geq 0 \), so \( A \geq G \) |
| Sum of a 7, 77, 777, …-type series (technique) | \( S_n = \frac{7}{9}\left(\frac{10(10^n-1)}{9} – n\right) \) |
| Fibonacci sequence (recurrence) | \( a_1 = a_2 = 1,\ a_n = a_{n-1} + a_{n-2} \) for \( n \gt 2 \) |
Sequences and Series Class 11 Formulas, Grouped by Topic
Sequences and Series
A sequence is a list of numbers in a definite order, generated by some rule; as a function, its domain is the set of natural numbers or a subset of them (NCERT, p. 137). The numbers are its terms \( a_1, a_2, a_3, \dots \), and \( a_n \) is the nth term or general term.
A series is the indicated sum of the terms of a sequence — the expression \( a_1 + a_2 + a_3 + \dots \), not the number it adds up to (NCERT, p. 137). In sigma notation, \[ \sum_{k=1}^{n} a_k = a_1 + a_2 + a_3 + \dots + a_n \]
The figure states the chapter’s organising idea: a sequence that follows a specific pattern is a progression. The geometric progression below is the pattern you will use most.
The Fibonacci sequence shows that not every sequence needs a direct formula — it is defined by the recurrence \( a_1 = a_2 = 1 \) and \( a_n = a_{n-1} + a_{n-2} \) for \( n \gt 2 \).
Geometric Progression (G.P.)
A sequence \( a_1, a_2, a_3, \dots \) is a geometric progression only when every term is non-zero and each term bears a constant ratio to the one before it (NCERT, p. 139):
\[ \frac{a_{k+1}}{a_k} = r \quad \text{for every } k \geq 1 \]
The constant \( r \) is the common ratio. Checking two or three ratios may reveal the pattern, but you confirm a G.P. only when all consecutive ratios agree. With first term \( a_1 = a \), the G.P. is \( a, ar, ar^2, ar^3, \dots \).
The general term of a G.P. is (NCERT, p. 140):
\[ a_n = ar^{n-1} \]
Why \( n-1 \)? Each new term multiplies the first term by \( r \) one more time, so the nth term carries \( r^{n-1} \). The formula also works for negative ratios, such as \( r = -\frac{1}{3} \).
Sum of First n Terms of a G.P.
Let \( S_n \) stand for the sum \( a + ar + ar^2 + \dots + ar^{n-1} \) (NCERT, p. 140; summary, NCERT, p. 149).
When \( r = 1 \), every term equals \( a \), so \[ S_n = na \]
When \( r \neq 1 \), \[ S_n = \frac{a(r^n – 1)}{r – 1} = \frac{a(1 – r^n)}{1 – r} \]
The two fraction forms are identical: multiplying top and bottom of the first by \( -1 \) gives the second. The \( r – 1 \) denominator is why \( r = 1 \) has to be handled separately. Inside the fraction, the exponent is always \( r^n \) — the sum runs over all n terms.
Sums like \( 7 + 77 + 777 + \dots \) are not G.P.s, but rewriting each term as a difference of powers of 10 reduces them to one (NCERT, p. 143):
\[ S_n = \frac{7}{9}\left[ \frac{10(10^n – 1)}{9} – n \right] \]
Geometric Mean (G.M.)
The geometric mean of two positive numbers \( a \) and \( b \) is (NCERT, p. 143):
\[ G = \sqrt{ab} \]
The three numbers \( a, G, b \) are consecutive terms of a G.P. To insert \( n \) geometric means \( G_1, G_2, \dots, G_n \) between \( a \) and \( b \), remember that \( b \) is the \( (n+2) \)th term of the new G.P. (NCERT, p. 143):
\[ b = ar^{n+1} \quad \Rightarrow \quad r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}} \]
\[ G_k = ar^k = a\left(\frac{b}{a}\right)^{\frac{k}{n+1}}, \quad k = 1, 2, \dots, n \]
The ratio can be negative too: inserting three numbers between 1 and 256 gives \( r = \pm 4 \), producing 4, 16, 64 or −4, 16, −64 (NCERT, p. 144).
Relationship Between A.M. and G.M.
For two positive real numbers \( a \) and \( b \), the arithmetic mean and the geometric mean are (NCERT, p. 144):
\[ A = \frac{a+b}{2}, \qquad G = \sqrt{ab} \]
Their difference is a perfect square divided by 2, so it is never negative (NCERT, p. 144):
\[ A – G = \frac{(\sqrt{a} – \sqrt{b})^2}{2} \geq 0 \quad \Rightarrow \quad A \geq G \]
Equality holds only when \( a = b \). This gives a built-in check: for two unequal positive numbers, the arithmetic mean is always the larger one.
What Each Symbol Means
All quantities are real numbers; the unit column below states the nature of each quantity, since pure mathematics has no SI unit.
| Symbol | What it means | Nature / unit |
|---|---|---|
| \( a \) | First term of a G.P.; also one of the two numbers in A.M./G.M. results | Real number |
| \( r \) | Common ratio, the constant value of \( a_{k+1}/a_k \) | Real number (ratio); \( r \neq 1 \) in the fraction sum formulas; \( r \neq 0 \) because G.P. terms are non-zero |
| \( n \) | Position of a term, or the number of terms being added | Positive integer (count) |
| \( a_n \) | The nth term (general term) of the sequence | Real number |
| \( S_n \) | Sum of the first \( n \) terms of a G.P. | Real number |
| \( l \) | Last term of a finite G.P. (notation used in the chapter) | Real number |
| \( A \) | Arithmetic mean of two numbers | Real number |
| \( G \) | Geometric mean of two positive numbers | Real number (positive) |
| \( G_1, G_2, \dots, G_n \) | The n geometric means inserted between \( a \) and \( b \) | Real numbers |
| \( b \) | Second number in A.M./G.M. results; the final number when means are inserted | Real number (positive in G.M. results) |
| \( k \) | Counting index in sigma notation and in \( a_{k+1} \) | Positive integer |
| \( \sum \) | Sigma: \( \sum_{k=1}^{n} a_k \) means the sum of \( a_k \) from \( k = 1 \) to \( k = n \) | Operator |
When to Use Each Formula
Before applying any G.P. formula, confirm the sequence really is a G.P.: the ratio \( a_{k+1}/a_k \) must be the same for every \( k \geq 1 \).
| Formula | Reach for it when … | Valid when |
|---|---|---|
| \( a_n = ar^{n-1} \) | You need a term deep in the G.P. without listing earlier terms, or must find which position holds a given value. | \( n \geq 1 \), sequence is a G.P. |
| \( S_n = \frac{a(1-r^n)}{1-r} \) | Summing the first \( n \) terms of a G.P. with a ratio between -1 and 1; keeps numerator and denominator positive. | \( r \neq 1 \) |
| \( S_n = \frac{a(r^n-1)}{r-1} \) | Summing the first \( n \) terms when \( r \gt 1 \); keeps both numerator and denominator positive. | \( r \neq 1 \) |
| \( S_n = na \) | All terms are equal (common ratio 1). | \( r = 1 \) |
| \( G = \sqrt{ab} \) | Finding the single geometric mean of two positive numbers. | \( a, b \) positive |
| \( r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}} \) | Inserting \( n \) numbers between \( a \) and \( b \) so the whole list is a G.P. | \( a \neq 0 \); \( b \) is the \( (n+2) \)th term |
| \( A = \frac{a+b}{2} \) | Finding the arithmetic mean of two numbers. | Any real \( a, b \) |
| \( A \geq G \) | Comparing the two means or checking answers in A.M.–G.M. problems. | \( a, b \) positive |
| \( \frac{7}{9}\left[\frac{10(10^n-1)}{9} – n\right] \) | Sums like \( 7 + 77 + 777 + \dots \) and scaled versions such as \( 8 + 88 + 888 + \dots \). | Rewrite \( 7, 77, 777, \dots \) as \( \frac{7}{9}(10^k – 1) \) |
Worked Examples
Example 1: Find the 8th term of the G.P. 3, 6, 12, 24, …
- Step 1: Identify the first term and common ratio: \( a = 3 \) and \( r = \frac{6}{3} = 2 \).
- Step 2: Apply the general-term formula \( a_n = ar^{n-1} \) with \( n = 8 \).
\[ a_8 = 3 \times 2^{8-1} = 3 \times 2^7 = 3 \times 128 = 384 \]
Final answer: The 8th term is \( 384 \).
Example 2: Which term of the G.P. 2, 6, 18, … is 1458?
Step 1: Here \( a = 2 \) and \( r = 3 \).
Set the general term equal to 1458 and solve for \( n \).
\[ 2 \times 3^{n-1} = 1458 \quad \Rightarrow \quad 3^{n-1} = 729 = 3^6 \]
Step 2: Equate the exponents: \( n – 1 = 6 \), so \( n = 7 \).
Final answer: 1458 is the 7th term. Check: \( 2 \times 3^6 = 1458 \).
Example 3: Find the sum of the first 6 terms of the G.P. 4, 2, 1, …
- Step 1: Here \( a = 4 \) and \( r = \frac{2}{4} = \frac{1}{2} \neq 1 \), so a fraction form of \( S_n \) applies.
- Step 2: Since \( |r| \lt 1 \), use \( S_n = \frac{a(1-r^n)}{1-r} \) with \( n = 6 \).
\[ S_6 = \frac{4\left(1 – \left(\frac{1}{2}\right)^6\right)}{1 – \frac{1}{2}} = \frac{4\left(1 – \frac{1}{64}\right)}{\frac{1}{2}} = 8 \times \frac{63}{64} = \frac{63}{8} \]
Final answer: \( S_6 = \frac{63}{8} = 7.875 \).
One-line A.M.–G.M. check: the numbers 18 and 8 have \( A = \frac{18+8}{2} = 13 \) and \( G = \sqrt{18 \times 8} = 12 \), and \( 13 \geq 12 \), exactly as \( A \geq G \) requires.
For practice on the textbook’s own questions, work through Exercises 8.1 and 8.2, then use the Maths Formulas index to jump to other chapters.
Common Mistakes to Avoid
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Using \( S_n = \frac{a(r^n-1)}{r-1} \) when \( r = 1 \), which divides by zero. | When \( r = 1 \), the series is \( a + a + \dots + a \), so \( S_n = na \). | If all terms are equal, the sum must be \( n \) times that term. |
| Writing the general term with the wrong exponent: \( a_n = ar^n \). | The nth term is \( a_n = ar^{n-1} \); the first term is \( a \), not \( ar \). | Put \( n = 1 \): the formula must return \( a \). |
| Putting \( r^{n-1} \) in the sum formula instead of \( r^n \). | The sum of n terms of a G.P. contains \( r^n \). | Put \( n = 1 \): \( S_1 \) must equal \( a \). |
| Using \( r = \left(\frac{b}{a}\right)^{\frac{1}{n}} \) when inserting n means between \( a \) and \( b \). | \( b \) is the \( (n+2) \)th term, so \( r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}} \). | Multiply \( a \) by \( r \) exactly \( n+1 \) times: you must get \( b \). |
| Confusing A.M. with G.M.: using \( \frac{a+b}{2} \) for the geometric mean. | \( A = \frac{a+b}{2} \) and \( G = \sqrt{ab} \); for positive \( a \neq b \), \( A \gt G \). | Test with 18 and 8: \( A = 13 \), \( G = 12 \), so \( A \geq G \) holds. |
Frequently Asked Questions
What is the difference between a sequence and a series?
A sequence is an ordered list of terms \( a_1, a_2, a_3, \dots \); it behaves like a function whose domain is the natural numbers. A series is the indicated sum of those terms, written \( a_1 + a_2 + a_3 + \dots \) or \( \sum_{k=1}^{n} a_k \) (NCERT, p. 137).
When do I use Sn = na instead of the fraction formula?
Only when the common ratio is \( r = 1 \), which means every term is the same number \( a \). Then \( S_n = a + a + \dots + a = na \). For every \( r \neq 1 \), use one of the two equivalent fraction forms.
Which form of the G.P. sum formula should I pick?
Both forms give the same number. Use \( S_n = \frac{a(1-r^n)}{1-r} \) when \( |r| \lt 1 \) and \( S_n = \frac{a(r^n-1)}{r-1} \) when \( r \gt 1 \), so that numerator and denominator stay positive. Either is valid whenever \( r \neq 1 \).
Why is the exponent 1/(n+1) when I insert n geometric means between a and b?
Because the last number \( b \) is the \( (n+2) \)th term of the new G.P., so \( b = ar^{n+1} \). Solving gives \( r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}} \). For example, inserting 2 means between 2 and 54 gives \( r = \left(\frac{54}{2}\right)^{\frac{1}{3}} = 3 \), producing 2, 6, 18, 54.
Reference: NCERT Class 11 Mathematics textbook, chapter Sequences and Series.
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