Let nϵN.If (1+x)n=a0x+a1x+a2x2+...+anxn, and an−3,an−2,an−1, are in AP then
A
a1,a2,a3 are in AP
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B
a1,a2,a3 are in HP
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C
n=7
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D
n=14
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Solution
The correct options are Aa1,a2,a3 are in AP Cn=7 an−3,an−2,an−1 are in A.P indicates that a3,a2,a1 are in A.P. Since they are binomial coefficients. For the above terms to be in A.P, they must follow the relationship (n−2r)2=n+2 Here r will be 2. Therefore n2−8r+16=n+2 n2−9r+14=0 (n−7)(n−2)=0 n=7 since n=2 is not possible here.