If log21/2 a + log21/4 a + log21/6 a+log21/8 a + ... upto 20 terms is 840, then a is equal to

Question

If log21/2 a + log21/4 a + log21/6 a+log21/8 a + ... upto 20 terms is 840, then a is equal to (A) 2 (B) 1 (C) 4 (D) √2

Tardigrade Admin · Accepted Answer

Given expression can be written as 2log2a + 4log2a + 6log2a + ... + 40log2a = log2a 2 + 4 + ... + 40 = 10(2 + 40) log2a = 420 log2a Now, 420 log2a = 840 ⇒ log2a = 2 ⇒ a = 4.

Q. If $l o g_{2^{1 / 2}} a + l o g_{2^{1 / 4}} a + l o g_{2^{1 / 6}} a + l o g_{2^{1 / 8}} a + ...$ upto $20$ terms is $840$ , then $a$ is equal to

$2$

$1$

$4$

$2$

Given expression can be written as
$2 l o g_{2} a + 4 l o g_{2} a + 6 l o g_{2} a + ... + 40 l o g_{2} a$
$= l o g_{2} a {2 + 4 + ... + 40}$
$= 10 (2 + 40) l o g_{2} a$
$= 420 l o g_{2} a$
Now, $420 l o g_{2} a = 840$
$\Rightarrow l o g_{2} a = 2$
$\Rightarrow a = 4$ .

Q. If log21/2​a+log21/4​a+log21/6​a+log21/8​a+... upto 20 terms is 840, then a is equal to

2

1

4

2​

Given expression can be written as 2log2​a+4log2​a+6log2​a+...+40log2​a =log2​a{2+4+...+40} =10(2+40)log2​a =420log2​a Now, 420log2​a=840 ⇒log2​a=2 ⇒a=4.

Q. If $l o g_{2^{1 / 2}} a + l o g_{2^{1 / 4}} a + l o g_{2^{1 / 6}} a + l o g_{2^{1 / 8}} a + ...$ upto $20$ terms is $840$ , then $a$ is equal to

$2$

$1$

$4$

$2$

Given expression can be written as
$2 l o g_{2} a + 4 l o g_{2} a + 6 l o g_{2} a + ... + 40 l o g_{2} a$
$= l o g_{2} a {2 + 4 + ... + 40}$
$= 10 (2 + 40) l o g_{2} a$
$= 420 l o g_{2} a$
Now, $420 l o g_{2} a = 840$
$\Rightarrow l o g_{2} a = 2$
$\Rightarrow a = 4$ .