Question

Calculate 1 - 2 - 3 - 4 - 5 - 6 +........+ 179 - 180​

Answer 1

Answer:

this can be splited into two Arithmetic progression:

1+3+5...+179 and -2-4-6...-180

Now, we will find sum of them separately using summation formula of AP

S_{n}S

n

= n/2{2a+(n-1)d}

where,S_{n}S

n

= sum of n terms

d = common difference

a = first term of AP

First, we will no of terms i.e. n of both APs

a) 1+3+5...+179 179 = a+(n-1)d

179 = 1+(n-1)2

179-1 = (n-1)2

178/2 = n-1

89+1 = n

n= 90

S_{n}S

n

(a)= n/2{2a+(n-1)d}

= 90/2{2*1 +(90-1)2}

= 45{2+89*2}

=45{2+178}

= 45*180 = 8100

b) -2-4-6...-180 = -(2+4+6...+180)

180 = a+(n-1)d

180 = 2+(n-1)2

180-2 = (n-1)2

178/2 = n-1

89+1 = n

n= 90

S_{n}S

n

(b)= n/2{2a+(n-1)d}

= 90/2{2*2 +(90-1)2}

= 45{4+89*2}

=45{4+178}

= 45*182

= 8190

Now, 1+3+5...+179-2-4-6...-180

S_{n}(a)S

n

(a) +S_{n}(b)S

n

(b) = 8100 - 8190 = -90

Step-by-step explanation:

however the answer is 90

Answer 2

S =   1 - 2 + 3 - 4 + 5 - 6 + ___ + 179 - 180

+   S =       1 - 2 + 3 -  4 + 5 - 6 + ___ + 179 - 180

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2S =           1 - 1 - 1 - 1 - 1 - 1 - 1 - 1 - ___ - 1 - 1 - 1 - 180

2S = -180

S =   $\frac{-180}{2}$

S = -90