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JEE Main 2020
Sequences & Series
Sequences and Series
Hard

Question

If logea,loge b,logec\log _e \mathrm{a}, \log _e \mathrm{~b}, \log _e \mathrm{c} are in an A.P. and logealoge2 b,loge2 bloge3c,loge3cloge\log _e \mathrm{a}-\log _e 2 \mathrm{~b}, \log _e 2 \mathrm{~b}-\log _e 3 \mathrm{c}, \log _e 3 \mathrm{c} -\log _e a are also in an A.P, then a:b:ca: b: c is equal to

Options

Solution

Key Concepts and Formulas

  1. Arithmetic Progression (A.P.) Property: If three numbers X,Y,ZX, Y, Z are in an A.P., then 2Y=X+Z2Y = X + Z.
  2. Logarithm Properties:
    • logbMlogbN=logb(MN)\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right)
    • klogbM=logb(Mk)k \log_b M = \log_b (M^k)
    • If logbM=logbN\log_b M = \log_b N, then M=NM = N.
  3. Domain of Logarithms: For logbX\log_b X to be defined, X>0X > 0.

Step-by-Step Solution

Step 1: Interpret the first A.P. condition and derive a relationship between a,b,ca, b, c.

We are given that logea,logeb,logec\log_e a, \log_e b, \log_e c are in an A.P. Using the A.P. property, twice the middle term is equal to the sum of the other two terms: 2logeb=logea+logec2 \log_e b = \log_e a + \log_e c Using the logarithm property klogbM=logb(Mk)k \log_b M = \log_b (M^k) on the left side and logbM+logbN=logb(MN)\log_b M + \log_b N = \log_b (MN) on the right side, we get: loge(b2)=loge(ac)\log_e (b^2) = \log_e (ac) Since the logarithms have the same base, their arguments must be equal: b2=ac()b^2 = ac \quad (*) This implies that a,b,ca, b, c themselves form a Geometric Progression (G.P.).

Step 2: Interpret the second A.P. condition and simplify the terms.

We are given that logealoge2b\log_e a - \log_e 2b, loge2bloge3c\log_e 2b - \log_e 3c, loge3clogea\log_e 3c - \log_e a are in an A.P. Let's simplify each term using the logarithm property logbMlogbN=logb(MN)\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right): Term 1: logealoge2b=loge(a2b)\log_e a - \log_e 2b = \log_e \left(\frac{a}{2b}\right) Term 2: loge2bloge3c=loge(2b3c)\log_e 2b - \log_e 3c = \log_e \left(\frac{2b}{3c}\right) Term 3: loge3clogea=loge(3ca)\log_e 3c - \log_e a = \log_e \left(\frac{3c}{a}\right)

Step 3: Apply the A.P. property to the simplified terms and derive a second relationship.

Since the three simplified terms are in an A.P., twice the middle term equals the sum of the first and the third term: 2(loge(2b3c))=loge(a2b)+loge(3ca)2 \left( \log_e \left(\frac{2b}{3c}\right) \right) = \log_e \left(\frac{a}{2b}\right) + \log_e \left(\frac{3c}{a}\right) Using the logarithm property klogbM=logb(Mk)k \log_b M = \log_b (M^k) on the left side: loge((2b3c)2)=loge(4b29c2)\log_e \left(\left(\frac{2b}{3c}\right)^2\right) = \log_e \left(\frac{4b^2}{9c^2}\right) Using the logarithm property logbM+logbN=logb(MN)\log_b M + \log_b N = \log_b (MN) on the right side: loge(a2b)+loge(3ca)=loge(a2b3ca)=loge(3c2b)\log_e \left(\frac{a}{2b}\right) + \log_e \left(\frac{3c}{a}\right) = \log_e \left(\frac{a}{2b} \cdot \frac{3c}{a}\right) = \log_e \left(\frac{3c}{2b}\right) Equating the arguments of the logarithms: 4b29c2=3c2b\frac{4b^2}{9c^2} = \frac{3c}{2b}

Step 4: Solve the derived equations to find the ratio a:b:ca:b:c.

From Step 3, we have: 4b29c2=3c2b\frac{4b^2}{9c^2} = \frac{3c}{2b} Cross-multiply: 4b22b=9c23c4b^2 \cdot 2b = 9c^2 \cdot 3c 8b3=27c38b^3 = 27c^3 Taking the cube root of both sides: 2b=3c2b = 3c This gives us a relationship between bb and cc: b=32corc=23bb = \frac{3}{2}c \quad \text{or} \quad c = \frac{2}{3}b

Now, we use the relationship from Step 1, b2=acb^2 = ac, and substitute the expression for cc: b2=a(23b)b^2 = a \left(\frac{2}{3}b\right) Since bb must be positive for logeb\log_e b to be defined, we can divide both sides by bb: b=23ab = \frac{2}{3}a This gives us a relationship between aa and bb: a=32ba = \frac{3}{2}b

Step 5: Express a,b,ca, b, c in terms of a common variable and find the ratio.

We have the following relationships: a=32ba = \frac{3}{2}b c=23bc = \frac{2}{3}b

Let's express a,b,ca, b, c in terms of bb: a:b:c=32b:b:23ba : b : c = \frac{3}{2}b : b : \frac{2}{3}b To get integer ratios, we can multiply all parts by the least common multiple of the denominators (2 and 3), which is 6: a:b:c=(32b6):(b6):(23b6)a : b : c = \left(\frac{3}{2}b \cdot 6\right) : (b \cdot 6) : \left(\frac{2}{3}b \cdot 6\right) a:b:c=(3b3):6b:(2b2)a : b : c = (3b \cdot 3) : 6b : (2b \cdot 2) a:b:c=9b:6b:4ba : b : c = 9b : 6b : 4b Dividing by bb (since b>0b > 0), we get the ratio: a:b:c=9:6:4a : b : c = 9 : 6 : 4

Step 6: Match the ratio with the given options.

The ratio 9:6:49:6:4 corresponds to option (B).


Common Mistakes & Tips

  • Domain of Logarithms: Always remember that the arguments of logarithms must be positive. This implies a,b,c>0a, b, c > 0.
  • Algebraic Errors: Be careful with algebraic manipulations, especially when dealing with fractions and exponents.
  • Misapplication of Logarithm Rules: Ensure you are using the correct logarithm properties for addition, subtraction, and multiplication by a constant.
  • Checking the A.P. Condition: Double-check that you are correctly applying the 2Y=X+Z2Y = X + Z property to the terms of the A.P.

Summary

The problem involves two conditions where terms are in an Arithmetic Progression. The first condition, involving logarithms of a,b,ca, b, c, leads to the relationship b2=acb^2 = ac, indicating that a,b,ca, b, c form a Geometric Progression. The second condition, involving differences of logarithms, simplifies to a relationship between the arguments of these logarithms. By applying the A.P. property to these simplified logarithmic terms, we derive a second algebraic relationship between a,b,ca, b, c. Solving these two relationships simultaneously allows us to determine the ratio a:b:ca:b:c.

The final answer is \boxed{9: 6: 4}, which corresponds to option (B).

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