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QJEEM23Ma2adf: JEE Main 2023 Mathematics - Class 11 JEEM23Ma2adf Solution: Let and be in A.P., and and be in G.P. If the sum of first 20…

Let a, b, c>1, a^3, b^3 and c3c^3 be in A.P., and log⁡ab,log⁡ca\log _a b, \log _c a and log⁡bc\log _b c be in G.P. If the sum of first 20 terms of an A.P., whose first term is a+4b+c3\frac{a+4 b+c}{3} and the common difference is a−8b+c10\frac{a-8 b+c}{10} is −444-444 , then abca b c is equal to :

Multiple Choice Options

  • A 343
  • B 216
  • C 3438\frac{343}{8}
  • D 1258\frac{125}{8}

Step-by-Step Verified Solution

2b3=a3+c32{b^3} = {a^3} + {c^3}

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

⇒(log⁡a)3=(log⁡c)3\Rightarrow {(\log a)^3} = {(\log c)^3}

⇒log⁡a=log⁡c\Rightarrow \log a = \log c

⇒a=c\Rightarrow a = c

⇒a=b=c\Rightarrow a = b = c

T1=2a,d=−3a5{T_1} = 2a,d = - {{3a} \over 5}

S20=−444{S_{20}} = - 444

⇒202(2(2a)+(19)(−3a5))=−444\Rightarrow {{20} \over 2}\left( {2(2a) + (19)\left( { - {{3a} \over 5}} \right)} \right) = - 444

⇒10(20a−57a)5=−444\Rightarrow 10{{(20a - 57a)} \over 5} = - 444

⇒37a=222\Rightarrow 37a = 222

⇒a=6\Rightarrow a = 6

⇒abc=(6)3=216\Rightarrow abc = {(6)^3} = 216

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