edited by
5,538 views

8 Answers

Best answer
33 votes
33 votes
$\textbf{P}=1+3+5+7+\ldots +(2k-1)$
$\quad=(2-1)+(4-1)+(6-1)+(8-1)+\ldots+(2k-1)$
$\quad=(2+4+6+8+\ldots+2k)+(-1-1-1-1-1\ldots k \  \text{times})$
$\quad=\textbf{Q}+(-k)=\textbf{Q-k}$

Correct Answer: $A$
edited by
12 votes
12 votes
Substitute k=3 then we get p=9 and q=12  on verifying we get option A.
6 votes
6 votes
The odd series is 1 3 5 7 ... 2k-1

So, 1+(t1-1)2=2k-1

      or t1=k;

P = (k/2) [2x1+(k-1)2]=k^2

The even series is 2 4 6 8 10  ... 2k

So, 2+(t2-1)2=2k

      or t2=k;

Q = (k/2) [2x2+(k-1)2]=k^2+k
So P=Q-k  is answer...
edited by
2 votes
2 votes
Just Write –

P = 1 + 3 + 5 + …

Q = 2 + 4 + 6 + …

and take K = 3

so P = (1+3+5) = 9

and Q = (2+4+6) = 12

and equation 1 holds-

P = Q – K

9 = 12 – 3.

it take only 30 second question to solve  :)
Answer:

Related questions

26 votes
26 votes
2 answers
2
Kathleen asked Sep 12, 2014
13,662 views
If a class $B$ network on the Internet has a subnet mask of $255.255.248.0$, what is the maximum number of hosts per subnet?$1022$$1023$$2046$$2047$