# If the coefficient of second, third and fourth terms in the expansion of (1 + x)2n are in A.P. Show that 2n2 – 9n + 7 = 0. - Mathematics

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If the coefficient of second, third and fourth terms in the expansion of (1 + x)2n are in A.P. Show that 2n2 – 9n + 7 = 0.

#### Solution

Given expression = (1 + x )2n

Coefficient of second term = 2nC1

Coefficient of third term = 2nC2

And coefficient of fourth term = 2nC3

As the given condition

2nC1. 2nC2 and 2nC3 are in A.P.

2nC22nC1 = 2nC32nC2

⇒ 2 * ""^(2n)"C"_2 = ""^(2n)"C"_1 + ""^(2n)"C"_3

⇒ 2 * (2n!)/(2!(2n - 2)!) = (2n!)/((2n - 1)!) + (2n!)/(3!(2n - 3)!)

⇒ 2[(2n(2n - 1)(2n - 2)!)/(2 xx 1 xx (2n - 2)!)] = (2n(2n - 1)!)/((2n - 1)!) + (2n(2n - 1)(2n - 2)(2n - 3)!)/(3 xx 2 xx 1 xx (2n - 3)!)

⇒ n(2n – 1) = n + (n(2n - 1)(2n - 2))/6

⇒ 2n – 1 = 1 + ((2n - 1)(2n - 2))/6

⇒ 12n – 6 = 6 + 4n2 – 4n – 2n + 2

⇒ 12n – 12 = 4n2 – 6n + 2

⇒ 4n2 – 6n – 12n + 2 + 12 = 0

⇒ 4n2 – 18n + 14 = 0

⇒ 2n2 – 9n + 7 = 0

Hence proved.

Concept: Binomial Theorem for Positive Integral Indices
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#### APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 8 Binomial Theorem
Exercise | Q 10 | Page 143
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