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

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Solution

The correct option is D
9

${2.14}_{c_{r}} = 14_{c_{r - 1}} + 14_{c_{r + 1}} \Rightarrow \frac{2}{(14 - r) r} = \frac{1}{(15 - r) (14 - r)} + \frac{1}{(r + 1) r} \Rightarrow r^{2} - 14 r + 45 = 0 \Rightarrow r = 5 (o r) r = 9$

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If the coefficients of Tr,Tr+1,Tr+2 terms of (1+x)14 are in A.P., then r =

The correct option is D 9 2.14cr=14cr−1+14cr+1⇒2(14−r)r=1(15−r)(14−r)+1(r+1)r⇒r2−14r+45=0⇒r=5(or)r=9

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

The correct option is D
9

${2.14}_{c_{r}} = 14_{c_{r - 1}} + 14_{c_{r + 1}} \Rightarrow \frac{2}{(14 - r) r} = \frac{1}{(15 - r) (14 - r)} + \frac{1}{(r + 1) r} \Rightarrow r^{2} - 14 r + 45 = 0 \Rightarrow r = 5 (o r) r = 9$