Exams Joint Entrance Examination Mathematics Sequences and Series Question 84 Question 84 of 519 Medium MCQ
Let a , b , c > 1 , a 3 , b 3 a, b, c>1, a^3, b^3 a , b , c > 1 , a 3 , b 3 and c 3 c^3 c 3 be in A.P., and log a b , log c a \log _a b, \log _c a log a b , log c a and log b c \log _b c 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 \frac{a+4 b+c}{3} 3 a + 4 b + c and the common difference is a − 8 b + c 10 \frac{a-8 b+c}{10} 10 a − 8 b + c
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2 b 3 = a 3 + c 3 2{b}^{3}={a}^{3}+{c}^{3} 2 b 3 = a 3 + c 3
( log a log c ) 2 = ( log b log a ) ( log c log b ) {(\frac{\log a}{\log c})}^{2}=(\frac{\log b}{\log a})(\frac{\log c}{\log b}) ( l o g c l o g a ) 2 = ( l o g a l o g b ) ( l o g b l o g c )
⇒ ( log a ) 3 = ( log c ) 3 \Rightarrow (\log a{)}^{3}=(\log c{)}^{3} ⇒ ( log a ) 3 = ( log c ) 3
⇒ log a = log c \Rightarrow \log a=\log c ⇒ log a = log c
⇒ a = c \Rightarrow a=c ⇒ a = c
⇒ a = b = c \Rightarrow a=b=c ⇒ a = b = c
T 1 = 2 a , d = − 3 a 5 {T}_{1}=2a,d=-\frac{3a}{5} T 1 = 2 a , d = − 5 3 a
S 20 = − 444 {S}_{20}=-444 S 20 = − 444
⇒ 20 2 ( 2 ( 2 a ) + ( 19 ) ( − 3 a 5 ) ) = − 444 \Rightarrow \frac{20}{2}(2(2a)+(19)(-\frac{3a}{5}))=-444 ⇒ 2 20 ( 2 ( 2 a ) + ( 19 ) ( − 5 3 a )) = − 444
⇒ 10 ( 20 a − 57 a ) 5 = − 444 \Rightarrow 10\frac{(20a-57a)}{5}=-444 ⇒ 10 5 ( 20 a − 57 a ) = − 444
⇒ 37 a = 222 \Rightarrow 37a=222 ⇒ 37 a = 222
⇒ a = 6 \Rightarrow a=6 ⇒ a = 6
⇒ a b c = ( 6 ) 3 = 216 \Rightarrow abc=(6{)}^{3}=216 ⇒ ab c = ( 6 ) 3 = 216 Answer the question to unlock