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Rangpur Higher Secondary 2011/7

Posted: Wed Feb 02, 2011 9:07 pm
by BdMO
Problem 7:
If $A + B =1$, $B + C = 2$, $C + D = 3$, ..., $X + Y = 25$, $Y + Z = 26$, find $A – Z$.

Re: Rangpur Higher Secondary 2011/7

Posted: Wed Feb 02, 2011 11:21 pm
by leonardo shawon
$A+B=1$
$B+C=2$
$C+D=3$
. . .
. . .
$X+Y=25$
$Y+Z=26$
( biog ) we get
$A-Z$=$1-2-3-. . . .-25-26$
so $A-Z=1-(1+2+. . . .+25+26)+1$
the result is $-349$.

Re: Rangpur Higher Secondary 2011/7

Posted: Wed Feb 02, 2011 11:37 pm
by Mehfuj Zahir
Your formula is uncorrect.(A+B)-(B+C)=A-C
A-C+C-D=A-D
(A-D)+(D+E)=A-E
A-E+E+F=A+F
DO IT CONSECUTIVELY AND GET THE ANS -13

Re: Rangpur Higher Secondary 2011/7

Posted: Thu Feb 03, 2011 12:34 am
by Tahmid Hasan
let's name the eqs 1,2,3,......25.then $1-2+3......$
get the ans :D

Re: Rangpur Higher Secondary 2011/7

Posted: Thu Feb 03, 2011 1:20 am
by leonardo shawon
Mehfuj Zahir wrote:Your formula is uncorrect.(A+B)-(B+C)=A-C
A-C+C-D=A-D
(A-D)+(D+E)=A-E
A-E+E+F=A+F
DO IT CONSECUTIVELY AND GET THE ANS -13
u misunderstood (may be)[\B]
thats a sequence from A to Z.

Re: Rangpur Higher Secondary 2011/7

Posted: Thu Feb 03, 2011 1:23 am
by Mehfuj Zahir
By doing it at last u get A-z=-13

Re: Rangpur Higher Secondary 2011/7

Posted: Thu Feb 03, 2011 1:55 am
by leonardo shawon
the Left side of Equation is $A-z$ ok? So the right side is
$1-2-3-4. . . . . . .-25-26$
=$1-(2+3+4+5. . . .+25+26)$
=$1-(1+2+3+4. . . .+25+26)+1$
=$2-(27*13)$
=$2-351$
=$-349$
... There are 26 equations. When we substract 26 equation at a time the first and last variables will left on the left side and one right side only the value of first equation will have positive sign.
or use a CALCULATOR.

Re: Rangpur Higher Secondary 2011/7

Posted: Thu Feb 03, 2011 2:21 am
by Mehfuj Zahir
it will like 1-2+3-4+5

Re: Rangpur Higher Secondary 2011/7

Posted: Thu Feb 03, 2011 11:09 am
by leonardo shawon
i told to substract all the equation.

Re: Rangpur Higher Secondary 2011/7

Posted: Thu Feb 03, 2011 11:53 am
by Mehfuj Zahir
If u dp it u don't have A-z