Question

Hello everyone ☺☺ 50 points.
IIT Based Question.

Suppose n be an integer greater than 1 . Let $a_n = \frac{1}{ log_{n}(2002) }$ . Suppose b = a2 + a3 + a4 + a5 and c = a10 + a11 + a12 + a13 + a14. Then find the value of ( b - c ) .

Please solve this .

Answer 1

Answer :

→ -1.

step-by-step explanation :

Solution :-

See the attachment.

$\huge \orange{ \boxed{ \sf \therefore b - c = \boxed{ - 1}.}}$

Hence, it is solved.

Answer 2

$\star\:\:\:\bf\large\underline\blue{Question:-}$

• Solve the given IIT based Question.

$\star\:\:\:\bf\large\underline\blue{Solution:-}$

Given that,

$a_{n} = \frac{1}{ log_{n}(2002) }$

Now,

$b = a_{2} + a_{3} + ...+a_{5}$

$c = a_{10} + a_{11} + ....+a _{14}$

We have to find the value of b-c

Here,

$a_{n} = \frac{1}{ log_{n}(2002) } = log_{2002}(n)$

So,

$b = log_{2002}(2) + log_{2002}(3) + ..+ log_{2002}(5)$

$= log_{2002}(2 \times 3 \times 4 \times 5)$

And,

$c = log_{2002}(10) + log_{2002}(11) + .... +log_{2002}(14)$

$= log_{2002}(10 \times 11 \times ...15)$

So,

$b - c = log_{2002}(2 \times 3 \times 4 \times 5) - log_{ 2002 }(10 \times 11 \times 12 \times 13 \times 14)$

$= log_{2002}( \frac{2 \times 3 \times 4 \times 5}{10 \times 11 \times 12 \times 14} )$

$= log_{2002}( \frac{1}{11 \times 13 \times 14} )$

$= log_{2002}( \frac{1}{2002} )$

$= log_{2002}( {2002}^{ - 1} )$

$= - 1 \: \cancel{ log_{2002}(2002) }$

$= - 1$

$\star\:\:\:\bf\large\underline\blue{Answer:-}$

• $b - c =- 1$

$\star\:\:\:\bf\large\underline\blue{Formulae\:Used:-}$

• $log_{a}(b) = \frac{1}{ log_{b}(a) }$
• $log(x) - log(y) = log(xy)$
• $log(x) + log(y) = log(xy)$
• $log_{x}( {y}^{n} ) = n \: log_{x}(y)$
• $log_{x}(x) = 1 \: for \: x \neq1$