17 votes 17 votes 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 $P = Q - k$ $P = Q + k$ $P = Q$ $P = Q + 2k$ Combinatory gatecse-2008 combinatory easy summation + – Kathleen asked Sep 11, 2014 • edited Apr 24, 2021 by Lakshman Bhaiya Kathleen 6.0k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
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: $A$ Pranabesh Ghosh 1 answered Jan 17, 2017 • edited Apr 23, 2019 by Naveen Kumar 3 Pranabesh Ghosh 1 comment Share Follow See all 0 reply Please log in or register to add a comment.
12 votes 12 votes Substitute k=3 then we get p=9 and q=12 on verifying we get option A. kireeti answered Oct 26, 2014 kireeti comment Share Follow See all 5 Comments See all 5 5 Comments reply Arjun commented Oct 26, 2014 reply Follow Share That's true. P is adding the first k odd numbers and Q is adding the first k even numbers. Each even number is 1 greater than its corresponding odd number being added. So for k additions, Q will be P+k 14 votes 14 votes kumar_sanjay commented Dec 10, 2016 reply Follow Share but option c also satisfies given condition........................... checking for each k 0 votes 0 votes Brij Mohan Gupta commented Jun 25, 2017 reply Follow Share @sanjay how option c will true please explain? 0 votes 0 votes SHIV_KANNAUJ commented May 1, 2020 reply Follow Share 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 0 votes 0 votes Vijay Dubey commented Feb 2, 2021 reply Follow Share @SHIV_KANNAUJ LoL 1 votes 1 votes Please log in or register to add a comment.
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... Palash Nandi 1 answered Nov 12, 2014 • edited Dec 2, 2017 by Puja Mishra Palash Nandi 1 comment Share Follow See 1 comment See all 1 1 comment reply Warrior commented Sep 5, 2017 reply Follow Share Nice approach.Thanks :-) Just small correction needed."The odd series is 2 4 6 8 10 ... 2k ." 0 votes 0 votes Please log in or register to add a comment.
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 :) Surya_Dev Chaturvedi answered Jan 15, 2021 Surya_Dev Chaturvedi comment Share Follow See all 0 reply Please log in or register to add a comment.