NCERT solution Class 11 binomial theorem

NCERT Solutions

Class 11 Maths

Binomial Theorem

NCERT Questions

Ex.8.1 Q.1
Ex.8.1 Q.2
Ex.8.1 Q.3
Ex.8.1 Q.4
Ex.8.1 Q.5
Ex.8.1 Q.6
Ex.8.1 Q.7
Ex.8.1 Q.8
Ex.8.1 Q.9
Ex.8.1 Q.10
Ex.8.1 Q.11
Ex.8.1 Q.12
Ex.8.1 Q.13
Ex.8.1 Q.14
Ex.8.2 Q.1
Ex.8.2 Q.2
Ex.8.2 Q.3
Ex.8.2 Q.4
Ex.8.2 Q.5
Ex.8.2 Q.6
Ex.8.2 Q.7
Ex.8.2 Q.8
Ex.8.2 Q.9
Ex.8.2 Q.10
Ex.8.2 Q.11
Ex.8.2 Q.12
Ex.Misc Q.1
Ex.Misc.Q.2
Ex.Misc.Q.3
Ex.Misc.Q.4
Ex.Misc.Q.5
Ex.Misc.Q.6
Ex.Misc.Q.7
Ex.Misc.Q.8
Ex.Misc.Q.9
Ex.Misc.Q.10

Ex.Misc Q.1

Find a, b and n in the expansion of (a + b)ⁿ if the first three terms of the expansion are 729, 7290 and 30375, respectively.

Given that the first three terms of the expansion are 729, 7290 and 30375, respectively.

Now T₁ = ⁿC₀ × a^n-0 × b⁰ = 729

=> aⁿ = 729 ................ (1)

T₂ = ⁿC₁ × a^n-1 × b¹ = 7290

=> n × a^n-1 × b = 7290 .......(2)

T₃ = ⁿC₂ × a^n-2 × b² = 30375

=> {n (n - 1) ÷ 2} × a^n-2 × b² = 30375 .......(3)

Now equation2/equation1

n × a^n-1 × (b ÷ aⁿ) = 7290 ÷ 729

=> n × = 10 .......(4)

Now equation (3) ÷ equation (2)

{n (n - 1) ÷ 2} × a^n-2 × b² ÷ n × a^n-1 × b = 30375 ÷ 7290

=> b (n - 1) ÷ 2a = 30375 ÷ 7290

=> b (n - 1) ÷ a = (30375 × 2) ÷ 7290

=> n - = 60750 ÷ 7290

=> 10 - = 6075 ÷ 729

(60750 and 7290 is divided by 10)

=> 10 - =

(6075 and 729 is divided by 243)

=> 10 - =

=> (30 - 25) ÷ 3 =

=> = ba

=> = ................. (5)

Put this value in equation (4), we get

n × = 10

=> 5n = 30

=> n =

=> n = 6

Now put this value in equation (1), we get

a⁶ = 729

=> a⁶ = 3⁶

=> a = 3

Now from equation (5), we get

=> b = 5

So, the values of a, b and n are 3, 5 and 6 respectively.

Video solution:

Binomial Theorem Class 11 Maths NCERT Chapter 7 NCERT Solutions

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NCERT solutions of Binomial Theorem Class 11 Maths - all questions of all exercises are solved. Q. No. 1, 2, 3, and so on. Complete explanation and detailed solution makes them the best videos for Class 11 Maths Binomial Theorem NCERT solutions.