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If the coefficients of T_r, T_{r+1}, T_{r+2} terms of (1+x)^{14}  are in A.P., then r = 

Option: 1

6


Option: 2

7


Option: 3

8


Option: 4

9


Answers (1)

best_answer

T_r={ }^{14} C_{r-1} x^{r-1} ; T_{r+1}={ }^{14} C_r x^r ; \quad T_{r+2}={ }^{14} C_{r+1} x^{r+1}

By the given condition 2 \cdot{ }^{14} C_r={ }^{14} C_{r-1}+{ }^{14} C_{r+1}     ....(i)

\begin{aligned} & 2 \cdot \frac{14 !}{r !(14-r) !}=\frac{14 !}{(r-1) !(15-r) !}+\frac{14 !}{(r+1) !(13-r) !} \\ \Rightarrow & \frac{2}{r \cdot(r-1) ! \cdot(14-r) \cdot(13-r) !} \\ = & \frac{1}{(r-1) ! \cdot(15-r) \cdot(14-r) \cdot(13-r) !}+ \\ & \frac{1}{(r+1) r(r-1) !(13-r) !} \\ \Rightarrow & \frac{2}{r(14-r)}=\frac{1}{(15-r)(14-r)}+\frac{1}{(r+1) r} \\ \Rightarrow & \frac{1}{r(14-r)}-\frac{1}{(15-r)(14-r)}=\frac{1}{(r+1) r}-\frac{1}{r(14-r)} \\ \Rightarrow & \frac{(15-r)-r}{r(15-r)(14-r)}=\frac{(14-r)-(r+1)}{(r+1) r(14-r)} \\ \Rightarrow & 15 r+15-2 r^2-2 r=195-30 r-13 r+2 r^2 \\ \Rightarrow & 4 r^2-56 r+180=0 \Rightarrow r^2-14 r+45=0 \\ \Rightarrow & (r-5)(r-9)=0 \Rightarrow r=5,9 \end{aligned}

But 5 is not given.  Hence r = 9.
Trick : Put the value of r from options in equation (i),  only (d) satisfy it.

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sudhir.kumar

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