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8 Answers

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Assme k=3 so 2k=6

p-sum of all odd numbers from 1 to 2k so p=1+3+5=9

q-sum of all even numbers from 1 to 2k so q=2+4+6=12

now as we can see P=Q-K
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$\therefore$ Answer is Option $\LARGE A$

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Try to use substitution method here

Let k=2

 We know that 1<= i <= 2k

That is 1<=i<= 4

For P i is odd 

For Q i is even

Therefore P will be 1+3 => 4

Q will be 2+4 =>6

Now check options and substitute values of P , Q ,k in options

a) P = Q-k

4 = 6-2

4=4

b) P=Q+k

4 = 6+2

4 != 8

c) P=Q

4!= 6

d) P=Q+2k

4 = 6+2(2)

4!= 10

Only option a) is satisfying

Therefore correct answer is option A

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$$P = \sum_{i=1,3,\dots,2k-1} i$$

transform the summation by putting $i  = 2r-1 $
where lower bound is $2r-1=1 \rightarrow r=1$
and upper bound is $2r-1=2k-1 \rightarrow r=k$

so we get

$$P = \sum_{r=1,2\dots,k} 2r-1$$

similarly
$$Q = \sum_{i=2,4,\dots,2k} i$$
transform the summation by putting $i  = 2r$
where lower bound is $2r=2 \rightarrow r=1$
and upper bound is $2r=2k \rightarrow r=k$

so we get

$$Q = \sum_{r=1,2\dots,k} 2r$$

doing $$Q – P = \sum_{r=1,2\dots,k} 2r – (2r-1) = \sum_{r=1,2\dots,k}1 = k$$

so ans is A i.e $P = Q – k$
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