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Question
In the binomial expansion of (1 + x)n, the coefficients of the 5th, 6th and 7th terms are in AP. Find all values of n
Solution
Coefficient of T5, T6, T7, are in A.P.
∴ nC4, nC5, nC6, are in A.P.
⇒ `("n"("n" - 1)("n" - 2)("n"- 3))/(4*3*2*1)`,
`("n"("n" - 1)("" - 2)("n" - 3)("n" - 4))/(5*4*3*2*1)`,
`("n"("n" - 1)("n"- 2)("n" - 3)("n" - 4)("n" - 5))/(6*5*4*3*1)` are in A.P.
Muliplying each erm by `(4*3*2*1)/("n"("n" - 1)("n" - 2)("n" - 3))`, we get
⇒ `1, ("n" - 4)/5, (("n" - 4)("n" - 5))/(6*5)` are in A.P.
⇒ `1, ("n" - 4)/5, ("n"^2 - 9"n" + 20)/30` arein A.P.
∴ `("n" - 4)/5 - 1 = ("n"^2 - 9"n" + 20)/30 - ("n" - 4)/5`
⇒ `("n" - 9)/5 =("n"^2 15"n" + 44)/30`
⇒ 6(n – 9) = n2 – 15n + 44
⇒ n2 – 21n + 98 = 0
⇒ (n – 1)(n – 14) = 0
∴ n = 7, 14
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