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Question
If 3rd, 4th 5th and 6th terms in the expansion of (x + a)n be respectively a, b, c and d, prove that `(b^2 - ac)/(c^2 - bd) = (5a)/(3c)`.
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Solution
Given expansion is (x + a)n.
`"T"_3 = a = ^nC_2 x^(n - 2) a^2 = (n(n - 1))/2 x^(n - 2) a^2`
`"T"_4 = b = ^nC_3 x^(n - 3) a^3 = (n(n - 1)(n - 2))/6 x^(n - 3) a^3`
`"T"_5 = c = ^nC_4 x^(n - 4) a^4 = (n(n - 1)(n - 2)(n - 3))/24 x^(n - 4) a^4`
`"T"_6 = d = ^nC_5 x^(n - 5) a^5 = (n(n - 1)(n - 2)(n - 3)(n - 4))/120 x^(n - 5) a^5`
Now,
`("T"_4)/("T"_3) = b/a = [(n(n - 1)(n - 2))/6 × x^(n - 3) × a^3]/[(n(n - 1))/2 × x^(n - 2) × a^2] = (n - 2)/3 . a/x ...(1)`
`("T"_5)/("T"_4) = c/b = [(n(n - 1)(n - 2)(n - 3))/24 × x^(n - 4) × a^4]/[(n(n - 1)(n - 2))/6 × x^(n - 3) × a^3] = (n - 3)/4 . a/x ...(2)`
`("T"_6)/("T"_5) = d/c = [(n(n - 1)(n - 2)(n-3)(n-4))/120 × x^(n - 5) × a^5]/[(n(n - 1)(n - 2)(n - 3))/24 × x^(n - 4) × a^4] = (n - 4)/5 . a/x ...(3)`
Again, dividing (1) by (2) and (2) by (3), we get
`[("T"_4)/("T"_3)]/[("T"_5)/("T"_4)] = [b/a]/[c/b] = [(n - 2)/3 . a/x]/[(n - 3)/4 . a/x] = [4(n - 2)]/[3(n - 3)]`
⇒ `(b^2)/(ac) = [4(n - 2)]/[3(n - 3)] ...(4)`
and
`[("T"_5)/("T"_4)]/[("T"_6)/("T"_5)] = [c/b]/[d/c] = [(n - 3)/4 . a/x]/[(n - 4)/5 . a/x] = [5(n- 3)]/[4(n - 4)].`
⇒ `[c^2]/[bd] = [5(n- 3)]/[4(n - 4)] ...(5)`
Now subtact 1 from both sides of equation (4) and (5) as:
⇒ `(b^2)/(ac) - 1 = [4(n - 2)]/[3(n - 3)] - 1`
⇒ `(b^2 - ac)/(ac) = (n + 1)/(3(n - 3))` ...(6)
and
⇒ `[c^2]/[bd] - 1 = [5(n- 3)]/[4(n - 4)] - 1`
⇒ `[c^2 - bd]/[bd] = [(n + 1)]/[4(n - 4)]` ...(7)
Again, on dividing (6) by (7), we get
`[(b^2 - ac)/(ac)]/[[c^2 - bd]/[bd]] = [(n + 1)/(3(n - 3))]/[[(n + 1)]/[4(n - 4)]]`
`(b^2 - ac)/(c^2 - bd) × (bd)/(ac) = [4(n - 4)]/[3(n - 3)]` ...(8)
On multiplying (5) by (8),
`(b^2 − ac)/(c^2 − bd) × (bd)/(ac) × c^2/(bd) = [4(n − 4)]/[3(n − 3)] × [5(n − 3)]/[4(n − 4)]`
⇒ `(b^2 − ac)/(c^2 − bd).c/a = 5/3`
⇒ `(b^2 - ac)/(c^2 - bd) = (5a)/(3c)`.
Hence proved.
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