Question

$\red {11}$
$If \:log_{10} 250, \:log_{10} 500\\and \: x \:are \:in \:A.P\:then \:find \:x$​

Answer 1

Answer: 3

Step-by-step explanation:

Given that,

$\log_{10}250\,,\log_{10}500\,\text{and x are in AP}\,$

Since we know that if a , b , c are in AP

2b = a + c

Here,

$a=\log_{10}250\\\;\\b=\log_{10}500\\\;\\c=x$

Applying the above condition for being these numbers in AP,

$2\log_{10}500=\log_{10}250+x\\\;\\2\log_{10}500-\log_{10}250=x\\\;\\x=2\log_{10}500-\log_{10}250\\\;\\x=\log_{10}500^2-\log_{10}250\;\;\;\;\;[\because\,n\log{x}=\log{x^n}]\\\;\\x=\log_{10}\frac{500^2}{250}\;\;\;\;\;[\because\log{m}-\log{n}=\log\frac{m}{n}\\\;\\x=\log_{10}\frac{500\times500}{250}\\\;\\x=\log_{10}{1000}\\\;\\x=\log_{10}10^3\\\;\\x=3\log_{10}10\;\;\;\;[\because \log_aa=1]\\\;\\x=3\times1\\\;\\x=3$

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

${\sf{{\red{log_{10}250}} \: , \: {\blue{log_{10}500}} \: and \: {\green{x}} \: are \: in \: A.P}}$

As we know if they are in A.P Then,

2b = a + c........(1)

So here ,

$\sf{a \: = \: log_{10}(250) } \\ \sf{b \: = \: log_{10}(500)} \\ \sf{c \: = \: x}$

We have to find value of x

Substitute these values in (1)

$\rightarrow {\sf{2log_{10}500 \: = \: log_{10}250 \: + \: x}}$

$\rightarrow {\sf{2log_{10}500 \: - \: log_{10}250 \: = \: x}}$

$\rightarrow {\sf{x \: = \: log_{10}500^2 \: - \: log_{10}250}}$

$\rightarrow {\sf{x \: = \: log_{10}{\frac{500^2}{250}}}}$

$\rightarrow {\sf{x \: = \: log_{10} \frac{250000}{250}}}$

$\rightarrow {\sf{x \: = \: log_{10} 1000}}$

$\rightarrow {\sf{x \: = \: log_{10} 10^3}}$

$\rightarrow {\sf{x \: = \: 3log_{10}{10}}}$

$\rightarrow {\sf{x \: = \: 3(1)}}$

$\Huge \implies {\boxed{\boxed{\sf{x \: = \: 3}}}}$

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