If the 3rd, 4th, 5th and 6th terms in the expansion of $(x + a)^{n}$ be respectively a,b,c and d, prove that $\frac{b^{2} - a c}{c^{2} - b d} = \frac{5 a}{3 c}$ .

Open in App

Solution

$\frac{T_{r + 1}}{T_{r}} = \frac{n - r + 1}{r} \cdot \frac{α}{x} : T_{3} = a, T_{4} = b, T_{5} = c, T_{6} = d$
Putting r = 3, 4, 5 in the above , we get
$\frac{T_{4}}{T_{3}} = \frac{n - 2}{3} \cdot \frac{α}{x} = \frac{b}{a}$ ....(1)
$\frac{T_{5}}{T_{4}} = \frac{n - 3}{4} \cdot \frac{α}{x} = \frac{c}{b}$ ...(2)
$\frac{T_{6}}{T_{5}} = \frac{n - 4}{5} \cdot \frac{α}{x} = \frac{d}{c}$ .....(3)
now dividing(1) by 2 and (2) by 3 respectively, we get
$\frac{4}{3} \cdot \frac{n - 2}{n - 3} = \frac{b^{2}}{a c}$ ......(4)
$\frac{5}{4} \cdot \frac{n - 3}{n - 4} = \frac{c^{2}}{b d}$ .....(5)
Subtracting 1 from both sides of (4) and (5),
$\frac{1}{3} \cdot \frac{n + 1}{n - 3} = \frac{b^{2} - a c}{a c}$ ......(6)
$\frac{1}{4} \cdot \frac{n + 1}{n - 4} = \frac{c^{2} - b d}{b d}$ ........(7)
Dividing (6) and (7), we get
$\frac{4}{3} \cdot \frac{n - 4}{n - 3} = \frac{b^{2} - a c}{c^{2} - b d} \cdot \frac{b d}{a c}$
or $\frac{4}{3} \cdot \frac{5}{4} \frac{b d}{c^{2}} = \frac{b^{2} - a c}{c^{2} - b d} \cdot \frac{b d}{a c} .$ ....by[5]
$∴ \frac{b^{2} - a c}{c^{2} - b d} = \frac{5 a}{3 c}$
Hence proved.

Suggest Corrections

Similar questions

Q. If 3rd, 4th 5th and 6th terms in the expansion of (x + a)ⁿ be respectively a, b, c and d, prove that

\frac{b^{2} - a c}{c^{2} - b d} = \frac{5 a}{3 c}

If 3rd, 4th, 5th and 6th terms in the expansion of $(x + α)^{n}$ be respectively a,b,c and d. prove that $\frac{b^{2} - a c}{c^{2} - b d} = \frac{5 a}{3 c} .$

If a,b,c and d in any binomial expansion be the 6th, 7th, 8th and 9th terms respectively, then prove that $\frac{b^{2} - a c}{c^{2} - b d} = \frac{4 a}{3 c} .$

Q. 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} - a c}{c^{2} - b d} = \frac{4 a}{3 c}

Q. The 3rd, 4th, 5th terms in the expansion of

{(x + a)}^{n}

are respectively

84, 280

and

560

. Find the values of

x, a

and

n

Join BYJU'S Learning Program

If the 3rd, 4th, 5th and 6th terms in the expansion of (x+a)n be respectively a,b,c and d, prove that b2−acc2−bd=5a3c.

If the 3rd, 4th, 5th and 6th terms in the expansion of $(x + a)^{n}$ be respectively a,b,c and d, prove that $\frac{b^{2} - a c}{c^{2} - b d} = \frac{5 a}{3 c}$ .