JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Sequences and Series

Let

a, b, c > 1, a^{3}, b^{3}

and

c^{3}

be in A.P., and

lo g_{a} b

lo g_{c} a

and

lo g_{b} c

be in G.P. If the sum of first 20 terms of an A.P., whose first term is

\frac{a + 4 b + c}{3}

and the common difference is

\frac{a - 8 b + c}{10}

is -444 , then abc is equal to

Answer: (B)

Solution

Since

a^{3}, b^{3}, c^{3}

are in A.P.,

a^{3} + c^{3} = 2 b^{3}

. The G.P. condition on

lo g_{a} b, lo g_{c} a, lo g_{b} c

gives

(lo g_{c} a)^{2} = lo g_{a} b \cdot lo g_{b} c

, that is

(\frac{lo g a}{lo g c})^{2} = \frac{lo g b}{lo g a} \cdot \frac{lo g c}{lo g b} = \frac{lo g c}{lo g a}

, so

(lo g a)^{3} = (lo g c)^{3}

and therefore

a = c

. Then

2 a^{3} = 2 b^{3}

forces

a = b = c

. The auxiliary A.P. has first term

\frac{a + 4 a + a}{3} = 2 a

and common difference

\frac{a - 8 a + a}{10} = - \frac{3 a}{5}

, so

S_{20} = \frac{20}{2} [2 (2 a) + 19 (- \frac{3 a}{5})] = - 74 a

. Setting

- 74 a = - 444

yields

a = 6

, and hence

ab c = 6^{3} = 216

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option