If the coefficients of r^th , (r+ 1)^th and (r + 2)^th terms in the binomial expansion of (1 + y)^m are in A.P., then m and r satisfy the equation: - Mathematics MCQ AIEEE 2026 | ExamTest.live

mediumMCQAIEEE2026Mathematics

1 mark

If the coefficients of r^{th}, (r + 1)^{th} and (r + 2)^{th} terms in the binomial expansion of (1 + y)^{m} are in A.P., then m and r satisfy the equation:

Solution & Step-by-step Explanation

The coefficients of r^{th}, (r + 1)^{th}, (r + 2)^{th} terms in (1 + y)^{m} are^{m} C_{r - 1},^{m} C_{r},^{m} C_{r + 1} respectively.Given they are in A.P., so: 2 \cdot^{m} C_{r} =^{m} C_{r - 1} +^{m} C_{r + 1} Divide by^{m} C_{r} : 2 = \frac{^{m} C _{r - 1}}{^{m} C _{r}} + \frac{^{m} C _{r + 1}}{^{m} C _{r}} Using the property \frac{^{m} C _{k}}{^{m} C _{k - 1}} = \frac{m - k + 1}{k} : 2 = \frac{r}{m - r + 1} + \frac{m - r}{r + 1} 2 = \frac{r ( r + 1 ) + ( m - r ) ( m - r + 1 )}{( m - r + 1 ) ( r + 1 )} 2 (m - r + 1) (r + 1) = r^{2} + r + m^{2} - m r + m - m r + r^{2} - r 2 (m r + m - r^{2} - r + r + 1) = 2 r^{2} + m^{2} - 2 m r + m 2 m r + 2 m - 2 r^{2} + 2 = 2 r^{2} + m^{2} - 2 m r + m m^{2} - 4 m r - m + 4 r^{2} - 2 = 0 m^{2} - m (4 r + 1) + 4 r^{2} - 2 = 0

Try it yourself before checking the explanation above.

If the coefficients of r^{th}, (r + 1)^{th} and (r + 2)^{th} terms in the binomial expansion of (1 + y)^{m} are in A.P., then m and r satisfy the equation:

m^{2} - m (4 r - 1) + 4 r^{2} - 2 = 0

m^{2} - m (4 r + 1) + 4 r^{2} + 2 = 0

m^{2} - m (4 r + 1) + 4 r^{2} - 2 = 0

m^{2} - m (4 r - 1) + 4 r^{2} + 2 = 0

Practice with timed mock tests and track your performance across Mathematics.