Sequences and Series 3 Question 11

11. If a,b,c,d and p are distinct real numbers such that (a2+b2+c2)p22(ab+bc+cd)p

+(b2+c2+d2)0, then a,b,c,d

(a) are in AP

(b) are in GP

(c) are in HP

(d) satisfy ab=cd

(1987, 2M)

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Answer:

Correct Answer: 11. (b)

Solution:

  1. Here, (a2+b2+c2)p22(ab+bc+cd)p

(a2p22abp+b2)+(b2p22bcp+c2)+(c2p22cdp+d2)0(apb)2+(bpc)2+(cpd)20

+(b2+c2+d2)0

[since, sum of squares is never less than zero] Since, each of the squares is zero.

(apb)2=(bpc)2=(cpd)2=0p=ba=cb=dc

a,b,c,d are in GP.



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