If log _ea,log _eb,log _ec are in an A.P. andn log _ea - log _e2b,n log _e2b - log _e3c,log _e3c - log _ea are also in ann A.P., then a:b:c is equal to

(b) : We have, log _e a, log _e b, log _e c are in an A.P.nnnn => 2 log _e b=log _e a+log _e c => b^2=a c \ (i)nnnnAlso, log _e a-log _e 2 b, log _e 2 b-log _e 3 c, log _e 3 c-log _e a are in an A.P.nnnn => 2(log _e 2 b-log _e 3 c)=log _e a-log _e 2 b+log _e 3 c-log _e annnnnn => (4 b^2 / 9 c^2)=(3 c / 2 b) \ (i i)nnn nFrom (i) and (ii), we havennnna=(9 c^2 / 4 c) => a=(9 / 4) c Therefore a: b: c=(9 / 4): (3 / 2): 1=9: 6: 4nnn

If \log_{e}a, \log_{e}b, \log_{e}c are in an A.P. and ... | Sequence & Series (Mathematics)

(b) : We have,

\log _e a, \log _e b, \log _e c

are in an A.P.\n\n

\n\n\Rightarrow 2 \log _e b=\log _e a+\log _e c \Rightarrow b^2=a c \#(i)\n\n

\n\nAlso,

\log _e a-\log _e 2 b, \log _e 2 b-\log _e 3 c, \log _e 3 c-\log _e a

are in an A.P.\n\n

\n\n\Rightarrow 2\left(\log _e 2 b-\log _e 3 c\right)=\log _e a-\log _e 2 b+\log _e 3 c-\log _e a\n\n

\n\n

\n\n\Rightarrow \frac{4 b^2}{9 c^2}=\frac{3 c}{2 b} \#(i i)\n\n

\n \nFrom (i) and (ii), we have\n\n

\n\na=\frac{9 c^2}{4 c} \Rightarrow a=\frac{9}{4} c \therefore a: b: c=\frac{9}{4}: \frac{3}{2}: 1=9: 6: 4\n\n

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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For any odd integer n ? 1,n^{3} - (n - 1)^{3} + + ( - 1)^{n - 1}1^{3} =

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Let \frac{1}{x_{1}},\frac{1}{x_{2}},,\frac{1}{x_{n}}( x_{i} eq 0 .\ for . \ i = 1,2,,n ) be in A.P. such that x_{1} = 4 and x_{21} = 20. If n is the least positive integer for which x_{n} > 50, then \sum_{i = 1}^{n}\mspace{2mu}( \frac{1}{x_{i}} ) is equal to

Let 3,a,b,c be in A.P. and 3,a - 1,b + 1,c + 9 be in G.P. Then, the arithmetic mean of a,b and c is

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