If the coefficients of 5th, 6th and 7th terms in the expansion of (1+x)nare in A.P., then n =

A
7 only
B
14 only
C
7 or 14
D
none of these

Solution:

Coefficient of $T_{5} {is}^{n} C_{4}$ ,that of $T_{6} {is}^{n} C_{5}$ and that of $T_{7} {is}^{n} C_{6}$ . According to the condition, $2^{n} C_{5} =^{n} C_{4} +^{n} C_{6}$ .Hence,
$2 [\frac{n!}{(n - 5)! 5!}] = [\frac{n!}{(n - 4)! 4!} + \frac{n!}{(n - 6)! 6!}]$
or $2 [\frac{1}{(n - 5) 5}] = [\frac{1}{(n - 4) (n - 5)} + \frac{1}{6 \times 5}]$
After solving, we get n = 7 or 14.

If the coefficients of 5^th, 6^th and 7^th terms in the expansion of $(1 + x)^{n}$ are in A.P., then n =

Solution:

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