MathsCBSE Class 11 · Binomial Theorem

General Term and Middle Term of (a + b)^n: Class 11 Binomial Theorem

Updated 17 October 2026 · Fact-checked

The general term of (a + b)^n is T(r+1) = nCr a^(n−r) b^r. If n is even there is one middle term, T(n/2 + 1). If n is odd there are two middle terms, T((n+1)/2) and T((n+3)/2). To find a term independent of x or a given power of x, write the power of x in T(r+1), equate it to the target and solve for r.

Term number is r+1: n=6 has one middle term T4 (r=3); n=5 has two middle terms T3 and T4 (r=2 and r=3).

Illustration drawn for QuestionBuddy.

Understand General Term and Middle Term of a Binomial Expansion

Writing out a whole binomial expansion is wasteful when a question asks for one term only. The general term does that job. Since the expansion of $(a + b)^{n}$ is $\sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r}$ , the term that has $b^{r}$ is the $(r + 1)$ th term:
$ $T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}, r = 0, 1, \dots, n .$ $
Note the index shift: $r = 0$ gives the first term, so the 5th term uses $r = 4$ . Always put the signs inside $b$ : for $(x - 2 y)^{n}$ take $b = - 2 y$ , so the sign $(- 1)^{r}$ comes with it.

Middle terms. The expansion has $n + 1$ terms.

If $n$ is even, $n + 1$ is odd and exactly one term sits in the middle: $T_{\frac{n}{2} + 1} =^{n} C_{n /2} a^{n /2} b^{n /2}$ .
If $n$ is odd, $n + 1$ is even and there are two middle terms, $T_{\frac{n + 1}{2}}$ and $T_{\frac{n + 3}{2}}$ . Their coefficients are equal ( $^{n} C_{(n - 1) /2} =^{n} C_{(n + 1) /2}$ ) but the powers of $a$ and $b$ differ.

Counting from the end. The $k$ th term from the end of $(a + b)^{n}$ is the $k$ th term from the start of $(b + a)^{n}$ , that is, $T_{n - k + 2}$ of $(a + b)^{n}$ .

Finding a particular term. Substitute the given $a$ and $b$ into $T_{r + 1}$ , collect the powers of $x$ and simplify to something like $x^{p - q r}$ .

*Term independent of $x$ :* solve $p - q r = 0$ . If $r$ is not a whole number between $0$ and $n$ , no such term exists.
*Coefficient of $x^{m}$ :* solve $p - q r = m$ , then read off the numerical part.

Using the same idea for products. For an expression like $(1 + 2 x)^{6} (1 - x)^{7}$ , the coefficient of $x^{5}$ is the sum over all splits $k + j = 5$ of the coefficient of $x^{k}$ in the first factor times the coefficient of $x^{j}$ in the second.

Conditions on coefficients. If a question says the coefficients of certain terms are in A.P. or equal, write each as $^{n} C_{r}$ and use $\frac{^{n} C _{r}}{^{n} C _{r - 1}} = \frac{n - r + 1}{r}$ to reduce the algebra. The general and middle terms are standard Class 11 binomial theorem skills, and they are the base of the JEE-foundation questions on this chapter.

General Term and Middle Term of a Binomial Expansion: formulas to remember

General term: $T (r + 1) = n C r \times a^{(} n - r) \times b^{r}$; Counting starts at r = 0 for the first term.
Middle term, n even: $T (n /2 + 1) = n C (n /2) \times a^{(} n /2) \times b^{(} n /2)$; Exactly one middle term.
Middle terms, n odd: $T ((n + 1) /2) an d T ((n + 3) /2)$; Two middle terms with equal coefficients.
Term from the end: $k t h f r o m e n d o f (a + b)^{n} = T (n - k + 2)$; Same as kth from start of (b + a)^n.
Ratio of consecutive coefficients: $n C (r) / n C (r - 1) = (n - r + 1) / r$; Quick way to handle A.P. and ratio conditions.

How to solve General Term and Middle Term of a Binomial Expansion questions

Use these steps for finding a specific term, the middle term or the term independent of x.

1Write the expansion as $(a + b)^{n}$ and decide what $a$ and $b$ are, signs included.
2Write $T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}$ and substitute $a$ and $b$ fully, keeping numbers, signs and powers of $x$ separate.
3For a named term (the 5th, the 3rd from the end), convert to $r$ first.
4For a power of $x$ , collect the exponent of $x$ in terms of $r$ , set it equal to the target and solve for $r$ .
5Check that $r$ is a whole number with $0 \leq r \leq n$ ; otherwise the term does not exist.
6Put the value of $r$ back, compute the numerical coefficient with the sign, and state the term.

General Term and Middle Term of a Binomial Expansion in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

General term as a counting rule

Suppose a quiz app sends 8 questions, and for each question the answer is marked either Correct or Wrong. Treat each question as a factor $(C + W)$ , so the whole quiz is $(C + W)^{8}$ .

How many different answer patterns have exactly 3 Wrong answers? Which term of the expansion gives this?

Try it first, then show the solutionSolution

Patterns with exactly 3 Wrong answers correspond to the term with $W^{3}$ , that is $T_{4}$ (since $r = 3$ ).
$T_{4} =^{8} C_{3} C^{5} W^{3}$ , and its coefficient counts the patterns.
$^{8} C_{3} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56$ .

Answer: 56 patterns; the 4th term.

Memory hook: The exponent of b counts how many times you chose b; the coefficient counts how many ways.

Using a few terms to estimate growth

Suppose Rs 10,000 is invested at 2% per year, compounded yearly, for 6 years. The amount is $10000 (1.02)^{6}$ .

Estimate the amount using the first three terms of the expansion of $(1 + 0.02)^{6}$ .

Try it first, then show the solutionSolution

$(1 + 0.02)^{6} \approx 1 +^{6} C_{1} (0.02) +^{6} C_{2} (0.02)^{2}$ .
$= 1 + 6 (0.02) + 15 (0.0004) = 1 + 0.12 + 0.006 = 1.126$ .
Amount $\approx 10000 \times 1.126 = 11260$ .
The fourth term $^{6} C_{3} (0.02)^{3} = 0.00016$ is tiny, so the estimate is close (exact value is about Rs 11,261.62).

Answer: About Rs 11,260.

Memory hook: When b is small, the first few terms carry almost everything.

Tricks to remember General Term and Middle Term of a Binomial Expansion

Short memory aids. Say each one aloud once, then use it on a question the same day.

r is one less than the term number
The (r+1)th term has b^r. Say it aloud: fifth term, r = 4.
Use it when: Avoiding the commonest slip in every general-term question.
Middle term count
Number of terms is n + 1: odd count means one middle, even count means two.
Use it when: Deciding between one and two middle terms.
Exponent hunt
For a power of x, write the exponent of x as a straight line in r and solve it.
Use it when: Independent term and coefficient questions.
Neighbour ratio
nC(r) = nC(r−1) × (n − r + 1)/r, so you never recompute big factorials.
Use it when: A.P. and ratio conditions on coefficients.

General Term and Middle Term of a Binomial Expansion: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Single middle term · 2 marks · Very short answerBoard level

Find the middle term of $(2 x - \frac{y}{2})^{8}$ .

Try it first, then show the solutionSolution

$n = 8$ is even, so the middle term is $T_{5}$ , with $r = 4$ . (½ mark)
$T_{5} =^{8} C_{4} (2 x)^{4} (- \frac{y}{2})^{4}$ . (½ mark)
$= 70 \cdot 16 x^{4} \cdot \frac{y ^{4}}{16}$ . (½ mark)
$= 70 x^{4} y^{4}$ . (½ mark)

Answer: $70 x^{4} y^{4}$

Example 2: Two middle terms · 3 marks · Short answerBoard level

Find the middle terms of $(2 x - \frac{x ^{2}}{4})^{7}$ .

Try it first, then show the solutionSolution

$n = 7$ is odd, so the middle terms are $T_{4}$ and $T_{5}$ . (½ mark)
$T_{4} =^{7} C_{3} (2 x)^{4} (- \frac{x ^{2}}{4})^{3} = 35 \cdot 16 x^{4} \cdot (- \frac{x ^{6}}{64})$ . (1 mark)
$= - \frac{35}{4} x^{10}$ . (½ mark)
$T_{5} =^{7} C_{4} (2 x)^{3} (- \frac{x ^{2}}{4})^{4} = 35 \cdot 8 x^{3} \cdot \frac{x ^{8}}{256} = \frac{35}{32} x^{11}$ . (1 mark)

Answer: $- \frac{35}{4} x^{10}$ and $\frac{35}{32} x^{11}$

Example 3: Term independent of x · 3 marks · Short answerHarder board level

Find the term independent of $x$ in $(\frac{3 x ^{2}}{2} - \frac{1}{3 x})^{9}$ .

Try it first, then show the solutionSolution

$T_{r + 1} =^{9} C_{r} (\frac{3 x ^{2}}{2})^{9 - r} (- \frac{1}{3 x})^{r}$ . (1 mark)
Power of $x$ : $2 (9 - r) - r = 18 - 3 r$ . For no $x$ , $18 - 3 r = 0$ , so $r = 6$ . (1 mark)
$T_{7} =^{9} C_{6} (\frac{3}{2})^{3} (\frac{1}{3})^{6} (- 1)^{6} = 84 \cdot \frac{27}{8} \cdot \frac{1}{729}$ . (½ mark)
$= \frac{84}{8 \cdot 27} = \frac{7}{18}$ . (½ mark)

Answer: $\frac{7}{18}$ (the 7th term)

Example 4: Coefficient in a product · 3 marks · Short answerHarder board level

Find the coefficient of $x^{5}$ in $(1 + 2 x)^{6} (1 - x)^{7}$ .

Try it first, then show the solutionSolution

Coefficient of $x^{k}$ in $(1 + 2 x)^{6}$ is $^{6} C_{k} 2^{k}$ : $1, 12, 60, 160, 240, 192$ for $k = 0..5$ . (1 mark)
Coefficient of $x^{j}$ in $(1 - x)^{7}$ is $(- 1)^{j}^{7} C_{j}$ ; for $j = 5, 4, 3, 2, 1, 0$ : $- 21, 35, - 35, 21, - 7, 1$ . (1 mark)
Sum of products with $k + j = 5$ : $- 21 + 420 - 2100 + 3360 - 1680 + 192$ . (½ mark)
$= 171$ . (½ mark)

Answer: 171

Example 5: Coefficients in A.P. · 5 marks · Long answerHOTS / competency

In the expansion of $(1 + x)^{n}$ , the coefficients of the 5th, 6th and 7th terms are in A.P. Find all possible values of $n$ .

Try it first, then show the solutionSolution

Coefficients are $^{n} C_{4},^{n} C_{5},^{n} C_{6}$ , so $2^{n} C_{5} =^{n} C_{4} +^{n} C_{6}$ . (1 mark)
Divide by $^{n} C_{5}$ : $2 = \frac{^{n} C _{4}}{^{n} C _{5}} + \frac{^{n} C _{6}}{^{n} C _{5}} = \frac{5}{n - 4} + \frac{n - 5}{6}$ . (1 mark)
Multiply by $6 (n - 4)$ : $12 (n - 4) = 30 + (n - 5) (n - 4)$ . (1 mark)
$12 n - 48 = n^{2} - 9 n + 50$ , so $n^{2} - 21 n + 98 = 0$ . (1 mark)
$(n - 7) (n - 14) = 0$ , so $n = 7$ or $n = 14$ (both allow a 7th term; check: $^{7} C_{5} \cdot 2 = 42 =^{7} C_{4} +^{7} C_{6} = 35 + 7$ ). (1 mark)

Answer: $n = 7$ or $n = 14$

Where students lose marks in General Term and Middle Term of a Binomial Expansion

The slips examiners see most often, and what to write instead.

Mistake: Using r = 5 for the 5th term
The (r+1)th term has b^r, so the 5th term needs r = 4.
Wrong working
5th term of $(a + b)^{9}$ $=^{9} C_{5} a^{4} b^{5}$ .
Right working
5th term $=^{9} C_{4} a^{5} b^{4}$ .
Do this: Write "term number = r + 1" before every general-term question.
Usually costs: The whole answer.
Mistake: Dropping the sign of b
In (x − y)^n the sign belongs to b, so every odd r gives a negative term.
Wrong working
$T_{4}$ of $(2 x - 3 y)^{6}$ taken as $+ 4320 x^{3} y^{3}$ .
Right working
$T_{4} =^{6} C_{3} (2 x)^{3} (- 3 y)^{3} = - 4320 x^{3} y^{3}$ .
Do this: Write b with its sign in brackets: $b = (- 3 y)$ .
Usually costs: 1 to 2 marks.
Mistake: Giving one middle term when n is odd
n + 1 is even, so two terms share the middle.
Wrong working
Middle term of $(x + y)^{7}$ is $T_{4}$ only.
Right working
Middle terms are $T_{4}$ and $T_{5}$ .
Do this: Check the parity of n first.
Usually costs: Half the marks.
Mistake: Not checking that r is a valid index
The equation for r may give a fraction or a value above n.
Wrong working
Exponent $9 - 2 r = 0$ gives $r = 4.5$ and the student still writes a term.
Right working
$r$ must be a whole number from 0 to 9, so no independent term exists.
Do this: State the validity check as a line of working.
Usually costs: The conclusion mark.

Exam tips

Write the general term with $a$ and $b$ substituted in full before simplifying; the method marks sit there.
Keep the numerical coefficient and the power of $x$ in two separate lines to avoid arithmetic slips.
State "n is even/odd, so there is one/two middle term(s)" in the first line of a middle-term answer.
For A.P. or ratio conditions use $\frac{^{n} C _{r}}{^{n} C _{r - 1}} = \frac{n - r + 1}{r}$ instead of factorials.
General and middle terms are standard Class 11 binomial theorem skills, so practise them along with the rest of the chapter.

Practice questions on General Term and Middle Term of a Binomial Expansion

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The term independent of $x$ in $(x + \frac{1}{x})^{10}$ is:
(a) 210
(b) 252
(c) 120
(d) 45
Hint
Power of $x$ is $10 - 2 r$ .
Answer
(b). $10 - 2 r = 0$ gives $r = 5$ , and $^{10} C_{5} = 252$ .
Q2 · 1 mark · MCQHarder board level
The 4th term from the end in the expansion of $(2 x - \frac{1}{x})^{8}$ is:
(a) $- \frac{448}{x ^{2}}$
(b) $\frac{448}{x ^{2}}$
(c) $- \frac{56}{x ^{2}}$
(d) $- 448 x^{2}$
Hint
4th from end is $T_{n - 4 + 2} = T_{6}$ .
Answer
(a). $T_{6} =^{8} C_{5} (2 x)^{3} (- \frac{1}{x})^{5} = 56 \cdot 8 x^{3} \cdot (- x^{- 5}) = - 448 x^{- 2}$ .
Q3 · 3 marks · Short answerHarder board level
Find the coefficient of $x^{5}$ in $(2 x^{2} - \frac{3}{x})^{10}$ .
Hint
Exponent of $x$ is $2 (10 - r) - r$ .
Answer
$T_{r + 1} =^{10} C_{r} (2 x^{2})^{10 - r} (- 3/ x)^{r}$ ; exponent $20 - 3 r = 5$ gives $r = 5$ . Coefficient $=^{10} C_{5} \cdot 2^{5} \cdot (- 3)^{5} = 252 \cdot 32 \cdot (- 243) = - 1959552$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The expansion of $(x + \frac{1}{x})^{9}$ has no term independent of $x$ .
Reason (R): The power of $x$ in the general term is $9 - 2 r$ , which is an odd number for every whole number $r$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
An odd number can never be zero.
Answer
(a). $9 - 2 r$ is odd, so it is never 0; hence no independent term.
Q5 · 3 marks · Short answerHOTS / competency
In the expansion of $(1 + x)^{43}$ , the coefficients of the $(2 r + 1)$ th and $(r + 2)$ th terms are equal. Find $r$ .
Hint
Equal $^{n} C_{a} =^{n} C_{b}$ means $a = b$ or $a + b = n$ .
Answer
Coefficients are $^{43} C_{2 r}$ and $^{43} C_{r + 1}$ . The question means two different terms, so $2 r \neq = r + 1$ and we need $2 r + r + 1 = 43$ , giving $r = 14$ (the 29th and 16th terms). $r = 1$ would give the same term twice, so it is rejected. So $r = 14$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the (r+1)th term of (a + b)^n?Show answer
nCr a^(n−r) b^r.
2.How many middle terms if n is 11?Show answer
Two: T6 and T7.
3.Which term is the kth from the end of (a + b)^n?Show answer
T(n − k + 2).
4.When does a term independent of x exist?Show answer
When the equation for r gives a whole number between 0 and n.
5.What is the ratio nCr / nC(r−1)?Show answer
(n − r + 1)/r.

General Term and Middle Term of a Binomial Expansion: frequently asked questions

Why does the first term use r = 0?

Because the first term has b^0, so the general term formula starts counting at r = 0 and the term number is always r + 1.

Is the middle term the same as the term with the greatest coefficient?

Yes for the binomial coefficient nCr: the largest value of nCr is at the middle term(s), but the actual term also carries powers of a and b, so the numerically greatest term can be elsewhere.

What if the question asks for the term from the end?

Convert it: the kth term from the end is T(n − k + 2) of (a + b)^n, then use the general term as usual.

Do I need these for the annual exam in 2026-27?

General and middle terms are standard Class 11 binomial theorem skills, so prepare them as part of the chapter. For the exact assessment scheme, follow the official CBSE syllabus.