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Question
If a, b, c and d in any binomial expansion be the 6th, 7th, 8th and 9th terms respectively, then prove that \[\frac{b^2 - ac}{c^2 - bd} = \frac{4a}{3c}\].
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Solution
Let the binomial expansion be (x + y)n
Given,
`T_6 = ^nC_5 x^(n - 5) y^5 = a`
`T_7 = ^nC_6 x^(n - 6) y^6 = b`
`T_8 = ^nC_7 x^(n - 7) y^7 = c`
`T_9 = ^nC_8 x^(n - 8) y^8 = d`
To prove: `(b^2 - ac)/(c^2 - bd)= (4a)/(3c)`
⇒ `(b^2 - ac)/(a) = 4/3 [(c^2 - bd)/(c)]`
⇒ `1/b [(b^2 - ac)/a] = 4/3 [(c^2 - bd)/(bc)]`
⇒ `b/a - c/b = 4/3 [c/d - d/c]` ...(i)
Now, substituting the values of a, b, c and d, we get
`(""^nC_6 x^(n - 6) y^6)/(""^nC_5 x^(n - 5) y^5) - (""^nC_7 x^(n - 7) y^7)/(""^nC_6 x^(n - 6) y^6) = 4/3 [(""^nC_7 x^(n - 7) y^7)/(""^nC_6 x^(n - 6) y^6) - (""^nC_8 x^(n -8) y^8)/(""^nC_7 x ^(n - 7) y^7)]`
`[(""^nC_6)/(""^nC_5) - (""^nC_4)/(""^nC_6)]y/x = 4/3 y/x [(""^nC_7)/(""^nC_6) - (""^nC_8)/(""^nC_7)]`
We know that, `(""^nC_r)/(""^nC_(r - 1)) = (n - r + 1)/r`
∴ `[(n - 5)/6 - (n - 6)/7] = 4/3[(n - 6)/7 - (n - 7)/8]`
⇒ `(7n - 35 - 6n + 36)/42 = (8n - 48 - 7n + 49)/(3 xx 7 xx 2)`
⇒ `(n + 1)/42 = (n + 1)/42`
⇒ LHS = RHS
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