tric and
(1+i)2011
[E-2011]
7.
(1-1)
2009 is equal to
1)-1 2) 1 3) 2
45-2
4)-2
Answers
Answered by
2
Answer:
Method 1:
It can solved as a combination of 2 APs.
AP1: 1+3+5+…45
The sum of 23 terms will be
S23 = (n/2)[2a+(n-1)d]
= (23/2)[2x1 + (23–1)2]
= (23/2)[2+22*2)
= 23[1+22]
= 23*23
= 529
AP2: -2–4–6…-44
The sum of 22 terms will be
S22 = (n/2)[2a+(n-1)d]
= (22/2)[2x(-2) + (22–1)(-2)]
= (22/2)[-4-21*2)
= 22[-2–21]
= 22*(-23)
= -506
AP1+AP2 = 529–506 = 23.
Answer = 23.
Method 2: The first and last number added = 46
The second and 44th number added = -46
So there are 22 pairs of numbers that cancel out each other and we are left with the middle term which is 23 as the sum of the whole series of 45 numbers.
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