In the expansion of ${(\frac{3}{2} + \frac{7}{3})}^{n}$ when $x = \frac{1}{2}$ if is known that $6^{t h}$ term is the greatest term the find the possible integral value of n.

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Solution

We know that ,

General term of ${(x + a)}^{n}$ is

$t_{r + 1} =^{n} C_{r} x^{n - r} . a^{r}$

Given that 6^th term is greatest in expansion of ${(\frac{3}{2} + \frac{7}{3})}^{n}$

So , 6^th term is

$t_{6} = t_{5 + 1} =^{n} C_{5} {(\frac{3}{2})}^{n - 5} \times {(\frac{7}{3})}^{5}$

Now

$n - 5 > 0$

$n > 5$

Hence value of n should be greater than 5 like 6

Hence n=6

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In the expansion of (32+73)n when x=12 if is known that 6th term is the greatest term the find the possible integral value of n.

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