If coefficient of $2^{n d}, 3^{r d}$ and $4^{t h}$ term in the expansion of ${(1 + x)}^{2 n}$ are in A.P. then :

2 n^{2} + 9 n + 7 = 0

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2 n^{2} - 9 n + 7 = 0

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2 n^{2} - 9 n - 7 = 0

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2 n^{2} + 9 n - 7 = 0

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Solution

The correct option is B $2 n^{2} - 9 n + 7 = 0$
$^{2 n} C_{1},^{2 n} C_{2},^{2 n} C_{3}$ are in A.P
$∴ 2 \times^{2 n} C_{2} =^{2 n} C_{1} +^{2 n} C_{3}$
$= 2 \times (\frac{2 n (2 n - 1)}{1 \times 2}) = \frac{2 n}{1} + \frac{2 n (2 n - 1) (2 n - 2)}{1 \times 2 \times 3}$
$= 2 n - 1 = 1 + \frac{(2 n - 1) (n - 1)}{3}$
$= 6 n - 3 = 3 n + 2 n^{2} - 3 n + 1$
$\Rightarrow 2 n^{2} - 9 n + 7 = 0$

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Similar questions

The coefficients of 2nd, 3rd and 4th terms in the expansion of $(1 + x)^{2 n}$ are in A.P., show that $2 n^{2} - 9 n + 7 = 0.$

2^{n d}

3^{r d}

4^{t h}

(1 + x)^{2 n}

2 n^{2} - 9 n + 7 = 0

2^{n d}, 3^{r d}, 4^{t h}

(1 + x)^{2 n}

A . P

2 n^{2} - 9 n + 8 =

2^{n d}, 3^{r d}

4^{t h}

(1 + x)^{2 n}

A . P

2 n^{2} - 9 n + 7

If the coefficients of 2nd, 3rd and 4th terms in the expansion of $(1 + x)^{2 n}$ are in AP, show that $2 n^{2} - 9 n + 7 = 0$

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