Question

calculate 1- 2 + 3 - 4 + 5 - 6 +____+179 - 180​

Answer 1

Answer:

The answer is -90

Step-by-step explanation:

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

Ans is -90

Answer 2

Step-by-step explanation:

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

Ans is -90