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Sequences and Series question

2020 · 6 Sep · Shift 1 · Q23
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Sequences and Series question

2020 · 6 Sep · Shift 1 · Q23

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
Let a , b, c , d and p be any non zero distinct real numbers such that (a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0. Then :
  1. A
    a, c, p are in G.P.
  2. B
    a, b, c, d are in G.P.
  3. C
    a, b, c, d are in A.P.
  4. D
    a, c, p are in A.P.
View written solutionFree

Correct answer: B

  1. Given quadratic in ppp

We are given

(a2+b2+c2)p2−2(ab+bc+cd)p+(b2+c2+d2)=0.(a^2+b^2+c^2)p^2-2(ab+bc+cd)p+(b^2+c^2+d^2)=0.(a2+b2+c2)p2−2(ab+bc+cd)p+(b2+c2+d2)=0.

Let

A=a2+b2+c2,B=ab+bc+cd,C=b2+c2+d2.A=a^2+b^2+c^2,\quad B=ab+bc+cd,\quad C=b^2+c^2+d^2.A=a2+b2+c2,B=ab+bc+cd,C=b2+c2+d2.

Then the equation is

Ap2−2Bp+C=0.Ap^2-2Bp+C=0.Ap2−2Bp+C=0.

Since a,b,c,d,pa,b,c,d,pa,b,c,d,p are real and ppp satisfies this quadratic, the discriminant must be non-negative:

(−2B)2−4AC≥0(-2B)^2-4AC \ge 0(−2B)2−4AC≥0

which gives

B2−AC≥0.B^2-AC \ge 0.B2−AC≥0.

So we study

(ab+bc+cd)2−(a2+b2+c2)(b2+c2+d2).(ab+bc+cd)^2-(a^2+b^2+c^2)(b^2+c^2+d^2).(ab+bc+cd)2−(a2+b2+c2)(b2+c2+d2).
  1. Apply Cauchy–Schwarz inequality

Consider the vectors

(a,b,c)and(b,c,d).(a,b,c) \quad \text{and} \quad (b,c,d).(a,b,c)and(b,c,d).

By Cauchy–Schwarz,

(ab+bc+cd)2≤(a2+b2+c2)(b2+c2+d2).(ab+bc+cd)^2 \le (a^2+b^2+c^2)(b^2+c^2+d^2).(ab+bc+cd)2≤(a2+b2+c2)(b2+c2+d2).

Hence,

B2−AC≤0.B^2-AC \le 0.B2−AC≤0.

But from the reality of ppp, we already need

B2−AC≥0.B^2-AC \ge 0.B2−AC≥0.

Therefore both together force

B2−AC=0.B^2-AC=0.B2−AC=0.

So equality holds in Cauchy–Schwarz.

  1. Equality condition in Cauchy–Schwarz

Equality in Cauchy–Schwarz holds iff the vectors are proportional:

(a,b,c)=λ(b,c,d)(a,b,c)=\lambda (b,c,d)(a,b,c)=λ(b,c,d)

for some real λ\lambdaλ. Thus,

a=λb,b=λc,c=λd.a=\lambda b,\quad b=\lambda c,\quad c=\lambda d.a=λb,b=λc,c=λd.

Therefore,

ab=bc=cd=λ.\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\lambda.ba​=cb​=dc​=λ.

So a,b,c,da,b,c,da,b,c,d are in geometric progression.

Hence Option B is true.

  1. Check the role of ppp

Since equality holds, we have

(ab+bc+cd)2=(a2+b2+c2)(b2+c2+d2).(ab+bc+cd)^2=(a^2+b^2+c^2)(b^2+c^2+d^2).(ab+bc+cd)2=(a2+b2+c2)(b2+c2+d2).

Therefore the quadratic has discriminant zero, so it has repeated root

p=ab+bc+cda2+b2+c2.p=\frac{ab+bc+cd}{a^2+b^2+c^2}.p=a2+b2+c2ab+bc+cd​.

Using

(a,b,c)=λ(b,c,d),(a,b,c)=\lambda (b,c,d),(a,b,c)=λ(b,c,d),

we get

ab+bc+cd=λ(b2+c2+d2),ab+bc+cd = \lambda (b^2+c^2+d^2),ab+bc+cd=λ(b2+c2+d2),

and also

a2+b2+c2=λ2(b2+c2+d2).a^2+b^2+c^2=\lambda^2(b^2+c^2+d^2).a2+b2+c2=λ2(b2+c2+d2).

Thus

p=λ(b2+c2+d2)λ2(b2+c2+d2)=1λ.p=\frac{\lambda (b^2+c^2+d^2)}{\lambda^2(b^2+c^2+d^2)}=\frac{1}{\lambda}.p=λ2(b2+c2+d2)λ(b2+c2+d2)​=λ1​.

But from the GP relation,

a=λ3d,c=λd,a=\lambda^3 d,\quad c=\lambda d,a=λ3d,c=λd,

so generally a,c,pa,c,pa,c,p need not be in GP or AP. Hence A and D are not necessarily true. Also GP does not imply AP, so C is false in general.

  1. Option-wise conclusion
  • A: False in general
  • B: True
  • C: False in general
  • D: False in general

Therefore the single correct answer is

B\boxed{\text{B}}B​
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