We know that (r + 1)^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n} is given by

T_{r+1} = ^{n}C_{r} a^{n-t} b^{r}

The first three terms of the expansion are given as 729, 7290 and 30375 respectively. Then we have,

T_{1} = ^{n}C_{0} a^{n-0} b^{0} = a^{n} = 729….. 1

T_{2} = ^{n}C_{1} a^{n-1} b^{1 }= na^{n-1} b = 7290…. 2

T_{3} = ^{n}C_{2} a^{n-2} b^{2} = {n (n -1)/2 }a^{n-2} b^{2}

= 30375……3

Dividing 2 by 1 we get

\( \dfrac{na^{n-1}b}{a^{n}} = \dfrac{7290}{729} \\ \dfrac{nb}{a} = 10\)

Dividing 3 by 2 we get

\( \dfrac{n(n-1)a^{n-2}b^{2}}{2na^{n-1}b} = \dfrac{30375}{7290} \\ \)

\( \dfrac{(n-1)b}{2a} = \dfrac{30375}{7290} \\ \dfrac{(n-1)b}{a} = \dfrac{30375}{7290} \times 2 \)

\( =\dfrac{25}{3} \\ \dfrac{(nb)}{a} -\dfrac{b}{a} \)

\( = \dfrac{25}{3} \\ \)

\( 10 -\dfrac{b}{a} = \dfrac{25}{3} \\ \)

\( \dfrac{b}{a} = 10 – \dfrac{25}{3} \)

\( = \dfrac{5}{3}\)

From 4 and 5 we have

n. \( \dfrac{5}{3}\) = 10

n = 6

Substituting n = 6 in 1 we get

a^{6} = 729

a = 3

From 5 we have, \( \dfrac{b}{3}\) = \( \dfrac{5}{3}\)

b = 5

Thus a = 3, b = 5 and n = 76

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