If A=1+3+5+.....+99,B=2+4+6+8+........+100 then the value of A-B is....?
Answers
n=?
a=1
l=99
d=2
l=a+(n-1)d
99=1+(n-1)2
n-1=98/2
n=50
A=50/2 (100)
A=2500
in B,
a=2
l=100
d=2
n=50
B=50/2 (102)
B=2550
A-B=2250-2550=-50
hope you will understand
Concept
The formula for finding the nth term of an Arithmetic progression (AP) is
an = a+(n-1)d
where an is the nth term
a is the 1st term
and d is the common difference
The formula for the sum of an AP is given by
Sum = n/2 [2a+(n-1)d]
Given
2 series A and B such that-
A=1+3+5+.....+99
B=2+4+6+8+......+100
Find
we need to find the value of A-B
Solution
We have,
A=1+3+5+.....+99
The above series is an AP with a= 1 and d=2
99 = 1 + 2 (n-1)
98/2 = (n-1)
49 = (n-1)
n = 50
Now, the sum of the AP will be
A = 50/2 [ 2 + (50-1) 2]
A = 25 [ 2+ 98 ]
A = 25 * 100
A = 2500
Similarly,
B=2+4+6+8+......+100
The above series is an AP with a= 2 and d= 2
100 = 2 + (n-1) 2
98/2 = (n-1)
49 = (n-1)
n = 50
Now, the sum of the AP will be
B = 50/2 [ (2*2) + (50-1) 2 ]
B = 25 (4 + 98)
B = 25 * 102
B = 2550
Therefore, A-B = 2500 -2550 = -50
Thus the value of A-B is -50
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