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If the 3rd, 4th. 5th and sixth term in t...

If the 3rd, 4th. 5th and sixth term in the expansion of `(x+alpha)^n` are a,b,c,d respectively, then prove that `((b^2-ac)/(c^2-bd))=(5a)/(3c).`

Text Solution

Verified by Experts

We know that
`(T_(r+1))/(T_(r)), (n-r+1)/(r)(alpha)/(x)`
`T_(3)=a,T_(4)=b,T_(5)=c,T_(6)=d`
Putting `r = 3, 4,5` in the abvoe we get
`(T_(4))/(T_(3)) = (n-2)/(3)(alpha)/(x) = b/a`
`(T_(5))/(T_(4)) = (n-3)/(4)(alpha)/(x) = c/b`
`(T_(6))/(T_(5)) = (n-4)/(5) (alpha)/(x) = d/c`
We have to prove
`(b^(2)-ac)/(c^(2)-bd) = (5a)/(3c)`
or `((b)/(c)-(a)/(b))/((c)/(b)-(d)/(c)) = (5a)/(3c)`
Now, `((b)/(c) - (a)/(b))/((c)/(b)-(d)/(c)) = ((4x)/((n-3)alpha)-(3x)/((n-2)alpha))/(((n-3)alpha)/(4x) - ((n-4)alpha)/(5x))`
`= (x^(2))/(alpha^(2))(4xx5)/((n-2)(n-3))(4n-8-3n+9)/(5n-15-4n+16)`
`= (x^(2))/(alpha^(2))(4xx5)/((n-2)(n-3))`
Also, `(5a)/(3c) = (x^(2))/(alpha^(2))(4xx5)/((n-2)(n-3))`
`:. L.H.S. = R.H.S.`
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