Let a, b, c>1, a³, b³ and c³ be in A.P., and log ₐ b, log _c a and log _b c be in G.P. If the sum of first 20 terms of an A.P., whose first term is (a+4 b+c)/3 and the common difference is (a-8 b+c)/10 is -444, then a b c is equal to :

Solution

Maths sir · 4 steps

Both conditions force a = b = c, so the A.P. depends on one number and the sum fixes it.

1. Use the G.P. The middle term squared equals the product of the outer two: (log_c a)² = logₐ b · log_b c = logₐ c. Put t = log_c a. Then logₐ c = 1/t, so t² = 1/t, which gives t³ = 1, so t = 1 and a = c.
2. Use the A.P. a³, b³, c³ in A.P. means 2b³ = a³ + c³ = 2a³. So b = a, and all three numbers are equal: a = b = c.
3. Build the new A.P. First term = (a + 4a + a)/3 = 2a. Common difference = (a - 8a + a)/10 = -3a/5. Now find the sum of 20 terms.

   S₂₀ = 20/2[2(2a) + 19(-3a/5)] = 10(4a - 57a/5) = -74a

4. So. -74a = -444 gives a = 6. Since a = b = c = 6, abc = 6³ = 216, which is option (b).

The idea to keep. When logs of a, b, c sit in a progression, the chain rule logₐ b · log_b c = logₐ c collapses them fast. Often the conditions force the numbers to be equal.

Answer from the NTA final answer key. Solution by PrepWiser.