Use the approach in Guass's Problems to find the following sum of arithmetic sequence a. 1+2+3+4...+101
b. 1+3+5+7...+997
c. 4+9+14+19...+494
d. 881+872+863+854+...+8
a. The sum of the sequence is

a) 5151 b) 249001 c)24651 d)43561

Step-by-step explanation:

a) a1=1 a2=2

d=a2-a1= 2-1=1

Define the quantity of the terms of this sequence

an=a1+d(n-1)

101= 1+ 1(n-1)

n=101

Sn= (a1+an)/2*n=( (1+101)/2)*101=51*101=5151

b) a1=1  a2=3

d=a2-a1=3-1=2

an=a1+d(n-1)

997= 1+ 2(n-1)

498= n-1

n=499

Sn= (a1+an)/2*n= (1+997)/2*499= 249001

c) a1=4 a2=9

d=a2-a1=9-4=5

an=a1+d(n-1)

494=4+5(n-1)

n=99

Sn= (a1+an)/2*n=( 4+494) /2*99= 498/2*99= 249*99= 24651

d) a1=881, a2=872

d=a2-a1= 872-881=-9

an= a1+d(n-1)

8=881-9*(n-1)

9(n-1)= 873

n=98

Sn= (a1+an)/2*n= (881+8)/2*98= 889*49= 43561

well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.