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Maths-

General

Easy

Question

If l, m, n are the $p to the power of text th end text end exponent comma q to the power of text th end text end exponent comma r to the power of text th end text end exponent$ terms of a G.P which are +ve, then $open vertical bar table attributes columnalign left end attributes row cell l o g invisible function application l end cell p 1 row cell l o g invisible function application m end cell q 1 row cell l o g invisible function application n end cell r 1 end table close vertical bar equals$

3
2
1
0

hint Hint:

The formula for the n^th term of the geometric progression is:
an = ar^n-1
where, a is the first term

r is the common ratio

n is the number of the term which we want to find.

The correct answer is: 0

Given : l, m, n are the $p to the power of text th end text end exponent comma q to the power of text th end text end exponent comma r to the power of text th end text end exponent$ terms of a G.P
We know that in GP
$a subscript n space equals space a r to the power of n minus 1 end exponent$

where,

a is the first term

r is the common ratio

n is the number of the term which we want to find.

Using this :
$p to the power of t h end exponent space t e r m space equals space l space equals space space equals space a r to the power of P minus 1 end exponent q to the power of t h end exponent space t e r m space equals space m space equals space space equals space a r to the power of Q minus 1 end exponent r to the power of t h end exponent space t e r m space equals space n space equals space space equals space a r to the power of R minus 1 end exponent$

Substituting these values in :
$rightwards double arrow open vertical bar table attributes columnalign left left left end attributes row cell log invisible function application l end cell p 1 row cell log invisible function application m end cell q 1 row cell log invisible function application n end cell r 1 end table close vertical bar equals open vertical bar table attributes columnalign left left left end attributes row cell log left parenthesis a r to the power of P minus 1 end exponent right parenthesis end cell p 1 row cell log left parenthesis a r to the power of Q minus 1 end exponent right parenthesis end cell q 1 row cell log left parenthesis a r to the power of R minus 1 end exponent right parenthesis end cell r 1 end table close vertical bar$

We know that
$rightwards double arrow log open parentheses a b close parentheses space equals space log a space plus log b space a n d rightwards double arrow log a to the power of m space equals space m log a$
Using these formulas and further simplifying
$rightwards double arrow open vertical bar table attributes columnalign left left left end attributes row cell log a space plus left parenthesis P minus 1 right parenthesis log space r right parenthesis end cell p 1 row cell log a space plus left parenthesis Q minus 1 right parenthesis log space r right parenthesis end cell q 1 row cell log a space plus space left parenthesis R minus 1 right parenthesis log space r right parenthesis end cell r 1 end table close vertical bar R subscript 2 space rightwards arrow R subscript 2 space minus space R subscript 1 space a n d space R subscript 3 rightwards arrow space R subscript 3 space minus space R subscript 1$
$rightwards double arrow open vertical bar table attributes columnalign left left left end attributes row cell log a space plus left parenthesis P minus 1 right parenthesis log space r right parenthesis end cell p 1 row cell left parenthesis Q minus P right parenthesis log space r end cell cell q space minus p end cell 0 row cell left parenthesis R minus P right parenthesis log space r end cell cell r space minus space p end cell 0 end table close vertical bar T a k i n g space o u t space c o m m o n space f r o m space R subscript 2 space a n d space R subscript 3 rightwards double arrow left parenthesis Q minus P right parenthesis left parenthesis R minus P right parenthesis open vertical bar table attributes columnalign left left left end attributes row cell log a space plus left parenthesis P minus 1 right parenthesis log space r right parenthesis end cell p 1 row cell log space r end cell 1 0 row cell log space r end cell 1 0 end table close vertical bar I f space 2 space r o w s space a r e space e q u a l space t h e n space t h e i r space d e t e r m i n a n t space i s space 0 rightwards double arrow left parenthesis Q minus P right parenthesis left parenthesis R minus P right parenthesis open vertical bar table attributes columnalign left left left end attributes row cell log a space plus left parenthesis P minus 1 right parenthesis log space r right parenthesis end cell p 1 row cell log space r end cell 1 0 row cell log space r end cell 1 0 end table close vertical bar space equals space 0$

Thus, $open vertical bar table attributes columnalign left end attributes row cell l o g invisible function application l end cell p 1 row cell l o g invisible function application m end cell q 1 row cell l o g invisible function application n end cell r 1 end table close vertical bar equals$ 0

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