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In the expansion of (1 + x)n,coefficients of 2nd ,3rd and 4th terms are in A.P.Then n is equal to
- a)7
- b)11
- c)9
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
In the expansion of (1 + x)n,coefficients of 2nd ,3rd and 4th terms ar...
Multiply both sides by 6 to eliminate the fraction:
6n(n-1) = 6n + n(n-1)(n-2)
Expanding both sides:
6n² - 6n = 6n + n³ - 3n² + 2n
Simplify:
6n² - 6n = n³ - 3n² + 8n
Rearrange terms:
n³ - 9n² + 14n = 0
Factor the equation:
n(n² - 9n + 14) = 0
We solve the quadratic equation n² - 9n + 14 = 0 using the quadratic formula:
n = (-(-9) ± √((-9)² - 4(1)(14)))/2(1)
n = (9 ± √(81 - 56))/2
n = (9 ± √25)/2
n = (9 ± 5)/2
n = (9 ± √(81 - 56))/2
n = (9 ± √25)/2
n = (9 ± 5)/2
So, the possible values of n are:
n = (9 + 5)/2 = 7 or n = (9 - 5)/2 = 2
Since n = 2 does not give the 2nd, 3rd, and 4th terms, the correct value of n is:
a) 7
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In the expansion of (1 + x)n,coefficients of 2nd ,3rd and 4th terms ar...
Given Information:
- Coefficients of 2nd, 3rd, and 4th terms in the expansion of (1 + x)n are in A.P.
Solution:
Step 1: Finding the general term of the expansion
The general term in the expansion of (1 + x)n is given by:
T(r + 1) = nCr * (1)r * (x)n-r
Step 2: Finding the coefficients
Let's find the coefficients of the 2nd, 3rd, and 4th terms using the general term:
- Coefficient of the 2nd term (r = 1): nC1 = n
- Coefficient of the 3rd term (r = 2): nC2 = n(n-1)/2
- Coefficient of the 4th term (r = 3): nC3 = n(n-1)(n-2)/6
Step 3: Using the information given
Given that the coefficients of the 2nd, 3rd, and 4th terms are in A.P:
n, n(n-1)/2, n(n-1)(n-2)/6 are in A.P
Step 4: Solving the A.P
To find n, we can solve the A.P. using the formula for an A.P:
2 * (n(n-1)/2) = n + n(n-1)(n-2)/6
n(n-1) = 2n + n(n-1)(n-2)/3
3n(n-1) = 6n + n(n-1)(n-2)
3n^2 - 3n = 6n + n^3 - 3n^2 - n
n^3 - 6n^2 + 9n = 0
n(n - 3)^2 = 0
Therefore, n = 0 or n = 3. Since n cannot be 0, n = 3 is the solution.
Conclusion:
Therefore, the value of n is 3, which is not one of the options provided.
- Coefficients of 2nd, 3rd, and 4th terms in the expansion of (1 + x)n are in A.P.
Solution:
Step 1: Finding the general term of the expansion
The general term in the expansion of (1 + x)n is given by:
T(r + 1) = nCr * (1)r * (x)n-r
Step 2: Finding the coefficients
Let's find the coefficients of the 2nd, 3rd, and 4th terms using the general term:
- Coefficient of the 2nd term (r = 1): nC1 = n
- Coefficient of the 3rd term (r = 2): nC2 = n(n-1)/2
- Coefficient of the 4th term (r = 3): nC3 = n(n-1)(n-2)/6
Step 3: Using the information given
Given that the coefficients of the 2nd, 3rd, and 4th terms are in A.P:
n, n(n-1)/2, n(n-1)(n-2)/6 are in A.P
Step 4: Solving the A.P
To find n, we can solve the A.P. using the formula for an A.P:
2 * (n(n-1)/2) = n + n(n-1)(n-2)/6
n(n-1) = 2n + n(n-1)(n-2)/3
3n(n-1) = 6n + n(n-1)(n-2)
3n^2 - 3n = 6n + n^3 - 3n^2 - n
n^3 - 6n^2 + 9n = 0
n(n - 3)^2 = 0
Therefore, n = 0 or n = 3. Since n cannot be 0, n = 3 is the solution.
Conclusion:
Therefore, the value of n is 3, which is not one of the options provided.
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In the expansion of (1 + x)n,coefficients of 2nd ,3rd and 4th terms are in A.P.Then n is equal toa)7b)11c)9d)none of theseCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In the expansion of (1 + x)n,coefficients of 2nd ,3rd and 4th terms are in A.P.Then n is equal toa)7b)11c)9d)none of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the expansion of (1 + x)n,coefficients of 2nd ,3rd and 4th terms are in A.P.Then n is equal toa)7b)11c)9d)none of theseCorrect answer is option 'A'. Can you explain this answer?.
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