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If rth ,(r+1)th and (r+2)th terms in the expansion of (1+x)n are in A.P. then
- a)(n+2r)3 = n−2
- b)(n−2r)2 = n+2
- c)(n+2r)2 = n+2
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
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If rth ,(r+1)th and (r+2)th terms in the expansion of(1+x)n are in A.P...
To determine the relationship between the terms (rth, (r+1)th, and (r+2)th) in the expansion of (1+x)^n, we can use the binomial theorem.
The general term of the expansion of (1+x)^n is given by:
C(n, r)*x^r*(1)^n-r
The rth term is given by:
C(n, r)*x^r*(1)^(n-r)
The (r+1)th term is given by:
C(n, r+1)*x^(r+1)*(1)^(n-r-1)
The (r+2)th term is given by:
C(n, r+2)*x^(r+2)*(1)^(n-r-2)
To determine if these terms are in an arithmetic progression (A.P.), we need to find the common difference between the terms. We can subtract consecutive terms to find the common difference:
(r+1)th term - rth term:
[C(n, r+1)*x^(r+1)*(1)^(n-r-1)] - [C(n, r)*x^r*(1)^(n-r)]
Simplifying this expression, we get:
C(n, r+1)*x^(r+1)*(1)^(n-r-1) - C(n, r)*x^r*(1)^(n-r)
= [C(n, r+1)*x^(r+1)*(1)^(n-r-1) - C(n, r)*x^r*(1)^(n-r)] / [x^r*(1)^(n-r)]
Similarly, the (r+2)th term - (r+1)th term can be found as:
[C(n, r+2)*x^(r+2)*(1)^(n-r-2)] - [C(n, r+1)*x^(r+1)*(1)^(n-r-1)]
Simplifying this expression, we get:
C(n, r+2)*x^(r+2)*(1)^(n-r-2) - C(n, r+1)*x^(r+1)*(1)^(n-r-1)
= [C(n, r+2)*x^(r+2)*(1)^(n-r-2) - C(n, r+1)*x^(r+1)*(1)^(n-r-1)] / [x^(r+1)*(1)^(n-r-1)]
Since we want the terms to be in an arithmetic progression, the common difference between these terms should be the same. Therefore, we can equate the expressions for the common difference and solve for x:
[C(n, r+1)*x^(r+1)*(1)^(n-r-1) - C(n, r)*x^r*(1)^(n-r)] / [x^r*(1)^(n-r)]
= [C(n, r+2)*x^(r+2)*(1)^(n-r-2) - C(n, r+1)*x^(r+1)*(1)^(n-r-1)] / [x^(r+1)*(1)^(n-r-1)]
Simplifying this expression and canceling out the common factors, we get:
[C(n, r+1)*x -
The general term of the expansion of (1+x)^n is given by:
C(n, r)*x^r*(1)^n-r
The rth term is given by:
C(n, r)*x^r*(1)^(n-r)
The (r+1)th term is given by:
C(n, r+1)*x^(r+1)*(1)^(n-r-1)
The (r+2)th term is given by:
C(n, r+2)*x^(r+2)*(1)^(n-r-2)
To determine if these terms are in an arithmetic progression (A.P.), we need to find the common difference between the terms. We can subtract consecutive terms to find the common difference:
(r+1)th term - rth term:
[C(n, r+1)*x^(r+1)*(1)^(n-r-1)] - [C(n, r)*x^r*(1)^(n-r)]
Simplifying this expression, we get:
C(n, r+1)*x^(r+1)*(1)^(n-r-1) - C(n, r)*x^r*(1)^(n-r)
= [C(n, r+1)*x^(r+1)*(1)^(n-r-1) - C(n, r)*x^r*(1)^(n-r)] / [x^r*(1)^(n-r)]
Similarly, the (r+2)th term - (r+1)th term can be found as:
[C(n, r+2)*x^(r+2)*(1)^(n-r-2)] - [C(n, r+1)*x^(r+1)*(1)^(n-r-1)]
Simplifying this expression, we get:
C(n, r+2)*x^(r+2)*(1)^(n-r-2) - C(n, r+1)*x^(r+1)*(1)^(n-r-1)
= [C(n, r+2)*x^(r+2)*(1)^(n-r-2) - C(n, r+1)*x^(r+1)*(1)^(n-r-1)] / [x^(r+1)*(1)^(n-r-1)]
Since we want the terms to be in an arithmetic progression, the common difference between these terms should be the same. Therefore, we can equate the expressions for the common difference and solve for x:
[C(n, r+1)*x^(r+1)*(1)^(n-r-1) - C(n, r)*x^r*(1)^(n-r)] / [x^r*(1)^(n-r)]
= [C(n, r+2)*x^(r+2)*(1)^(n-r-2) - C(n, r+1)*x^(r+1)*(1)^(n-r-1)] / [x^(r+1)*(1)^(n-r-1)]
Simplifying this expression and canceling out the common factors, we get:
[C(n, r+1)*x -
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If rth ,(r+1)th and (r+2)th terms in the expansion of(1+x)n are in A.P. thena)(n+2r)3= n−2b)(n−2r)2 = n+2c)(n+2r)2 = n+2d)none of theseCorrect answer is option 'B'. 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 If rth ,(r+1)th and (r+2)th terms in the expansion of(1+x)n are in A.P. thena)(n+2r)3= n−2b)(n−2r)2 = n+2c)(n+2r)2 = n+2d)none of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If rth ,(r+1)th and (r+2)th terms in the expansion of(1+x)n are in A.P. thena)(n+2r)3= n−2b)(n−2r)2 = n+2c)(n+2r)2 = n+2d)none of theseCorrect answer is option 'B'. Can you explain this answer?.
If rth ,(r+1)th and (r+2)th terms in the expansion of(1+x)n are in A.P. thena)(n+2r)3= n−2b)(n−2r)2 = n+2c)(n+2r)2 = n+2d)none of theseCorrect answer is option 'B'. 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 If rth ,(r+1)th and (r+2)th terms in the expansion of(1+x)n are in A.P. thena)(n+2r)3= n−2b)(n−2r)2 = n+2c)(n+2r)2 = n+2d)none of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If rth ,(r+1)th and (r+2)th terms in the expansion of(1+x)n are in A.P. thena)(n+2r)3= n−2b)(n−2r)2 = n+2c)(n+2r)2 = n+2d)none of theseCorrect answer is option 'B'. Can you explain this answer?.
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