JEE MainMathematics

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+4 marks−1 if incorrectNumericalPrevious-year question

Finding the Middle Coefficient Given Ratios of Consecutive Binomial Terms

Q: Why is there an extra factor of 2 in the denominator of (c_r)/(c_(r+1))?

Because the coefficient is binomnk2^k. Dividing c_r by c_(r+1) gives (2^r)/(2^(r+1)) = 1/2 alongside the ratio of combinations.

Let the coefficients of three consecutive terms in the binomial expansion of (1+2 x)^n be in the ratio 2: 5: 8. Then the coefficient of the term, which is in the middle of these three terms, is

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Step-by-step solutionView

Correct answer

The coefficient of the middle term among the three consecutive terms is 1120.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

In the expansion of $(1 + 2 x)^{n}$ , the coefficients of three consecutive terms $T_{r + 1}, T_{r + 2}, T_{r + 3}$ are in the ratio $2 : 5 : 8$ .

02Goal

Find the coefficient of the middle of these three terms, which is $T_{r + 2}$ .

03Approach

The general term is $T_{k + 1} = (\binom{n}{k}) 2^{k} x^{k}$ , with coefficient $c_{k} = (\binom{n}{k}) 2^{k}$ . Set up two ratios $\frac{c_{r}}{c_{r + 1}} = \frac{2}{5}$ and $\frac{c_{r + 1}}{c_{r + 2}} = \frac{5}{8}$ using the property $\frac{(\binom{n}{k})}{(\binom{n}{k + 1})} = \frac{k + 1}{n - k}$ , solve for $n$ and $r$ , then compute $c_{r + 1} = (\binom{n}{r + 1}) 2^{r + 1}$ .

04Execute

From the first ratio: $\frac{(\binom{n}{r}) 2^{r}}{(\binom{n}{r + 1}) 2^{r + 1}} = \frac{2}{5} ⟹ \frac{r + 1}{2 (n - r)} = \frac{2}{5} ⟹ \frac{r + 1}{n - r} = \frac{4}{5}$ $5 r + 5 = 4 n - 4 r ⟹ 4 n - 9 r = 5 — (1)$

05Execute

From the second ratio: $\frac{(\binom{n}{r + 1}) 2^{r + 1}}{(\binom{n}{r + 2}) 2^{r + 2}} = \frac{5}{8} ⟹ \frac{r + 2}{2 (n - r - 1)} = \frac{5}{8} ⟹ \frac{r + 2}{n - r - 1} = \frac{5}{4}$ $4 r + 8 = 5 n - 5 r - 5 ⟹ 5 n - 9 r = 13 — (2)$

06Execute

Subtracting (1) from (2): $(5 n - 9 r) - (4 n - 9 r) = 13 - 5 ⟹ n = 8$ Substitute $n = 8$ into (1): $4 (8) - 9 r = 5 ⟹ 32 - 5 = 9 r ⟹ r = 3$

07Execute

The middle term is $T_{r + 2} = T_{5}$ . Its coefficient is: $(\binom{n}{r + 1}) 2^{r + 1} = (\binom{8}{4}) 2^{4} = 70 \times 16 = 1120$

✓Verify

Verify the coefficients: $c_{3} = (\binom{8}{3}) 2^{3} = 56 \times 8 = 448$ , $c_{4} = 1120$ , $c_{5} = (\binom{8}{5}) 2^{5} = 56 \times 32 = 1792$ . The ratio is $448 : 1120 : 1792$ . Dividing each by $224$ gives $2 : 5 : 8$ , exactly matching the given ratio.

Hints that build this answer step by step

What is the general formula for the coefficient of the $r$ -th power of $x$ (which is the $(r + 1)$ -th term) in the expansion of $(1 + 2 x)^{n}$ ?
$(\binom{n}{r}) 2^{r}$
Using three consecutive coefficients $C_{r - 1}$ , $C_{r}$ , and $C_{r + 1}$ with ratio $2 : 5 : 8$ , which system of equations in $n$ and $r$ results from $\frac{C_{r}}{C_{r - 1}} = \frac{5}{2}$ and $\frac{C_{r + 1}}{C_{r}} = \frac{8}{5}$ ?
$4 (n - r + 1) = 5 r$ and $5 (n - r) = 4 (r + 1)$
What are the values of $n$ and $r$ obtained by solving the system $4 n - 9 r + 4 = 0$ and $5 n - 9 r - 4 = 0$ ?
$n = 8$ and $r = 4$
What is the numerical value of the middle term's coefficient $C_{r} = (\binom{8}{4}) 2^{4}$ ?
1120

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why is there an extra factor of 2 in the denominator of

\frac{c_{r}}{c_{r + 1}}

Because the coefficient is $(\binom{n}{k}) 2^{k}$ . Dividing $c_{r}$ by $c_{r + 1}$ gives $\frac{2^{r}}{2^{r + 1}} = \frac{1}{2}$ alongside the ratio of combinations.

Answer

Accepted answer: 1120

The coefficient of the middle term among the three consecutive terms is 1120.

Why each option works or fails

Step-by-step solution

Given: In the expansion of $(1 + 2 x)^{n}$ , the coefficients of three consecutive terms $T_{r + 1}, T_{r + 2}, T_{r + 3}$ are in the ratio $2 : 5 : 8$ .
Goal: Find the coefficient of the middle of these three terms, which is $T_{r + 2}$ .
Approach: The general term is $T_{k + 1} = (\binom{n}{k}) 2^{k} x^{k}$ , with coefficient $c_{k} = (\binom{n}{k}) 2^{k}$ . Set up two ratios $\frac{c_{r}}{c_{r + 1}} = \frac{2}{5}$ and $\frac{c_{r + 1}}{c_{r + 2}} = \frac{5}{8}$ using the property $\frac{(\binom{n}{k})}{(\binom{n}{k + 1})} = \frac{k + 1}{n - k}$ , solve for $n$ and $r$ , then compute $c_{r + 1} = (\binom{n}{r + 1}) 2^{r + 1}$ .
Execute: From the first ratio: $\frac{(\binom{n}{r}) 2^{r}}{(\binom{n}{r + 1}) 2^{r + 1}} = \frac{2}{5} ⟹ \frac{r + 1}{2 (n - r)} = \frac{2}{5} ⟹ \frac{r + 1}{n - r} = \frac{4}{5}$ $5 r + 5 = 4 n - 4 r ⟹ 4 n - 9 r = 5 — (1)$
Execute: From the second ratio: $\frac{(\binom{n}{r + 1}) 2^{r + 1}}{(\binom{n}{r + 2}) 2^{r + 2}} = \frac{5}{8} ⟹ \frac{r + 2}{2 (n - r - 1)} = \frac{5}{8} ⟹ \frac{r + 2}{n - r - 1} = \frac{5}{4}$ $4 r + 8 = 5 n - 5 r - 5 ⟹ 5 n - 9 r = 13 — (2)$
Execute: Subtracting (1) from (2): $(5 n - 9 r) - (4 n - 9 r) = 13 - 5 ⟹ n = 8$ Substitute $n = 8$ into (1): $4 (8) - 9 r = 5 ⟹ 32 - 5 = 9 r ⟹ r = 3$
Execute: The middle term is $T_{r + 2} = T_{5}$ . Its coefficient is: $(\binom{n}{r + 1}) 2^{r + 1} = (\binom{8}{4}) 2^{4} = 70 \times 16 = 1120$
Verify: Verify the coefficients: $c_{3} = (\binom{8}{3}) 2^{3} = 56 \times 8 = 448$ , $c_{4} = 1120$ , $c_{5} = (\binom{8}{5}) 2^{5} = 56 \times 32 = 1792$ . The ratio is $448 : 1120 : 1792$ . Dividing each by $224$ gives $2 : 5 : 8$ , exactly matching the given ratio.

Finding the Middle Coefficient Given Ratios of Consecutive Binomial Terms

What feels right?

The coefficient of the middle term among the three consecutive terms is 1120.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

07Execute

✓Verify

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