GATE CSE 2008 | Question: 24

Question

Let $P =\sum \limits_ {i\;\text{odd}}^{1\le i \le 2k} i$ and $Q = \sum\limits_{i\;\text{even}}^{1 \le i \le 2k} i$, where $k$ is a positive integer. Then

• A. $P = Q - k$
• B. $P = Q + k$
• C. $P = Q$
• D. $P = Q + 2k$

Answer 1

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

Correct Answer: $A$

Answer 2

Substitute k=3 then we get p=9 and q=12  on verifying we get option A.

Comments

@sanjay how option c will true please explain?

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Lets Assume the value of k=5

then we take number from 1 to 10.

then P= 1+3+5+7+9 = 25

and Q= 2+4+6+8+10 = 30            So, here we conclude that P=Q+K

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@SHIV_KANNAUJ

LoL

Answer 3

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...

Comments

Nice approach.Thanks :-)

Just small correction needed."The odd series is 2 4 6 8 10  ... 2k ." 

Answer 4

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  :)