Question

1,-2,-3,+2,-3,-4,+3,-4,-5+...find the sum of 99th term
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Answer 1

Answer:

The sum of given sequence for first 99 terms is  -660.

Step-by-step explanation:

We have 99 terms,

1 set of numbers are positive integers

the other 2 set of numbers are negative integers

so let’s divide 99 by 3 which will give us 3 times of 33

Now we have 3 times of 33.

So now we have 33 positive integers (the first set 1+2+3+4………33)

and rest i.e 99–33 = 66 negative integers

Now we know we have 2 sets of negative integers

So we must divide 66 / 2 = 33

NOW,

the first sequence is 1+2+3+4+………33

the second sequence is -2 -3 -4 -5 -…-34

the third sequence is -3 -4 -5 -6 -……….-35

let’s find the sum of first sequence

Here

a=1 (first number)

l= 33 (last number)

n= 33 (number of terms)

Therefore, S = n/2(a+l)

S= 33/2(1+33)

S=16.5(34)

S= 561

let’s find the sum of second sequence

Here

a=-2 (first number)

l= -34 (last number)

n= 33 (number of terms)

Therefore, S = n/2(a+l)

S= 33/2(-2 + (-34))

S=16.5(-2–34)

S=16.5(-36)

S= -594

let’s find the sum of third sequence

Here

a= -3 (first number)

l= -35(last number)

n= 33 (number of terms)

Therefore, S = n/2(a+l)

S= 33/2(-3 + (-35))

S=16.5(-3–35)

S=16.5(-38)

S= -627

Now adding all the three sums we get 561+(-594)+(-627) = -660

Answer 2

Answer:

8

Step-by-step explanation:

1-2=-1

-1+3=2

2+2=4

4-3=