[tex]\red {11}[/tex]
[tex]If \:log_{10} 250, \:log_{10} 500\\and \: x \:are \:in \:A.P\:then \:find \:x [/tex]
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[tex]\red {11}[/tex]
[tex]If \:log_{10} 250, \:log_{10} 500\\and \: x \:are \:in \:A.P\:then \:find \:x [/tex]
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Answer: 3
Step-by-step explanation:
Given that,
[tex]\log_{10}250\,,\log_{10}500\,\text{and x are in AP}\,[/tex]
Since we know that if a , b , c are in AP
2b = a + c
Here,
[tex]a=\log_{10}250\\\;\\b=\log_{10}500\\\;\\c=x[/tex]
Applying the above condition for being these numbers in AP,
[tex]2\log_{10}500=\log_{10}250+x\\\;\\2\log_{10}500-\log_{10}250=x\\\;\\x=2\log_{10}500-\log_{10}250\\\;\\x=\log_{10}500^2-\log_{10}250\;\;\;\;\;[\because\,n\log{x}=\log{x^n}]\\\;\\x=\log_{10}\frac{500^2}{250}\;\;\;\;\;[\because\log{m}-\log{n}=\log\frac{m}{n}\\\;\\x=\log_{10}\frac{500\times500}{250}\\\;\\x=\log_{10}{1000}\\\;\\x=\log_{10}10^3\\\;\\x=3\log_{10}10\;\;\;\;[\because \log_aa=1]\\\;\\x=3\times1\\\;\\x=3[/tex]
[tex]\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}[/tex]
[tex]{\sf{{\red{log_{10}250}} \: , \: {\blue{log_{10}500}} \: and \: {\green{x}} \: are \: in \: A.P}}[/tex]
As we know if they are in A.P Then,
2b = a + c........(1)
So here ,
[tex] \sf{a \: = \: log_{10}(250) } \\ \sf{b \: = \: log_{10}(500)} \\ \sf{c \: = \: x}[/tex]
We have to find value of x
Substitute these values in (1)
[tex] \rightarrow {\sf{2log_{10}500 \: = \: log_{10}250 \: + \: x}}[/tex]
[tex] \rightarrow {\sf{2log_{10}500 \: - \: log_{10}250 \: = \: x}}[/tex]
[tex] \rightarrow {\sf{x \: = \: log_{10}500^2 \: - \: log_{10}250}}[/tex]
[tex] \rightarrow {\sf{x \: = \: log_{10}{\frac{500^2}{250}}}}[/tex]
[tex] \rightarrow {\sf{x \: = \: log_{10} \frac{250000}{250}}}[/tex]
[tex] \rightarrow {\sf{x \: = \: log_{10} 1000}}[/tex]
[tex] \rightarrow {\sf{x \: = \: log_{10} 10^3}}[/tex]
[tex]\rightarrow {\sf{x \: = \: 3log_{10}{10}}}[/tex]
[tex]\rightarrow {\sf{x \: = \: 3(1)}}[/tex]
[tex]\Huge \implies {\boxed{\boxed{\sf{x \: = \: 3}}}}[/tex]
[tex]\rule{200}{2}[/tex]
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