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JEE Mathematics 2023 Question with Solution

Let a,b,c>1a, b, c > 1, a3,b3a^3, b^3 and c3c^3 be in A.P., and log⁡ba,log⁡ac\log_b a, \log_a c and log⁡cb\log_c b be in G.P. If the sum of the first 2020 terms of an A.P., whose first term is a+4b+c3\frac{a + 4b + c}{3} and the common difference is a−8b+c10=−444\frac{a - 8b + c}{10} = -444, then abcabc is equal to:

  • A

    343343

  • B

    216216

  • C

    343343

  • D

    1258\frac{125}{8}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: a3,b3,c3a^3, b^3, c^3 are in A.P. and log⁡ba,log⁡ac,log⁡cb\log_b a, \log_a c, \log_c b are in G.P.

Find: the value of abcabc.

Since a3,b3,c3a^3, b^3, c^3 are in A.P.,

a3+c3=2b3a^3 + c^3 = 2b^3

This is equation (1).

Also, log⁡ba,log⁡ac,log⁡cb\log_b a, \log_a c, \log_c b are in G.P., so

(log⁡ac)2=log⁡ba⋅log⁡cb\left(\log_a c\right)^2 = \log_b a \cdot \log_c b

Using change of base,

log⁡ba=log⁡alog⁡b,log⁡cb=log⁡blog⁡c,log⁡ac=log⁡clog⁡a\log_b a = \frac{\log a}{\log b}, \quad \log_c b = \frac{\log b}{\log c}, \quad \log_a c = \frac{\log c}{\log a}

Hence,

log⁡alog⁡b⋅log⁡blog⁡c=(log⁡clog⁡a)2\frac{\log a}{\log b} \cdot \frac{\log b}{\log c} = \left(\frac{\log c}{\log a}\right)^2

So,

log⁡alog⁡c=(log⁡clog⁡a)2\frac{\log a}{\log c} = \left(\frac{\log c}{\log a}\right)^2

which gives

(log⁡a)3=(log⁡c)3(\log a)^3 = (\log c)^3

Therefore,

a=ca = c

This is equation (2).

From (1) and (2),

a3+a3=2b3a^3 + a^3 = 2b^3

So,

2a3=2b3⇒a=b=c2a^3 = 2b^3 \Rightarrow a = b = c

Now the first term of the required A.P. is

T1=a+4b+c3=a+4a+a3=2aT_1 = \frac{a + 4b + c}{3} = \frac{a + 4a + a}{3} = 2a

And the common difference is

d=a−8b+c10=a−8a+a10=−6a10=−3a5d = \frac{a - 8b + c}{10} = \frac{a - 8a + a}{10} = \frac{-6a}{10} = -\frac{3a}{5}

Using the sum formula for an A.P.,

S20=202[2T1+19d]S_{20} = \frac{20}{2}\left[2T_1 + 19d\right]

Substituting T1=2aT_1 = 2a and d=−3a5d = -\frac{3a}{5},

S20=10[4a+19(−3a5)]S_{20} = 10\left[4a + 19\left(-\frac{3a}{5}\right)\right] =10[20a−57a5]= 10\left[\frac{20a - 57a}{5}\right] =−74a= -74a

Given that

S20=−444S_{20} = -444

therefore,

−74a=−444⇒a=6-74a = -444 \Rightarrow a = 6

Since a=b=c=6a = b = c = 6,

abc=63=216abc = 6^3 = 216

Therefore, the correct option is B.

Direct Symmetry Approach

Given: the two progression conditions connect a,b,ca, b, c.

Find: abcabc quickly.

From the G.P. condition on log⁡ba,log⁡ac,log⁡cb\log_b a, \log_a c, \log_c b, the solution shows this reduces to

a=ca = c

Then the A.P. condition on a3,b3,c3a^3, b^3, c^3 becomes

a3,b3,a3a^3, b^3, a^3

being in A.P., which is possible only when

a3=b3a^3 = b^3

Hence,

a=b=ca = b = c

because all are greater than 11.

So immediately,

T1=a+4b+c3=2a,d=a−8b+c10=−3a5T_1 = \frac{a + 4b + c}{3} = 2a, \quad d = \frac{a - 8b + c}{10} = -\frac{3a}{5}

Then

S20=202[2(2a)+19(−3a5)]=−74aS_{20} = \frac{20}{2}\left[2(2a) + 19\left(-\frac{3a}{5}\right)\right] = -74a

Given S20=−444S_{20} = -444,

a=6a = 6

Thus,

abc=63=216abc = 6^3 = 216

So the correct option is B.

Common mistakes

  • Assuming the G.P. condition means the product of the three logarithms is constant. That is incorrect; for three terms in G.P., the square of the middle term equals the product of the other two. Use (log⁡ac)2=log⁡ba⋅log⁡cb\left(\log_a c\right)^2 = \log_b a \cdot \log_c b.

  • Using the A.P. condition on a3,b3,c3a^3, b^3, c^3 as if it implied a,b,ca, b, c are in A.P. This is wrong because the terms in A.P. are the cubes, not the numbers themselves. Write a3+c3=2b3a^3 + c^3 = 2b^3 first.

  • Substituting into the A.P. sum formula with the wrong first term. Here T1=a+4b+c3T_1 = \frac{a + 4b + c}{3}, and after using a=b=ca=b=c it becomes 2a2a, not 3a+4b+c\frac{3}{a+4b+c} or any inverted form.

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