Question

without adding,find the sum of 1+3+5+7+9+11+13+15+17+19

Answer 1

Hello friend,
This is possible by arithmetic progression.
a=1
d=2
a(n)=19
a+(n-1)d=19
1+(n-1)2=19
n-1=18/2
n=10
S(n)=n[2a+(n-1)d]/2=10[2×1+(10-1)2]/2=5[2+18]=100
So, answer is 100.
hope it helps.

Answer 2

1+3+5+7+9+11+13+15+17+19=?

We have to find it using Arithmetic Progression.
a=1
d=2
tn=19

a+(n-1)d=19
1+(n-1)2=19
(n-1)2=19-1
(n-1)2=18
n-1=18/2
n-1=9
n=9+1
n=10

Sn=n[2a+(n-1)d]/2
Sn=10(2(1)+(10-1)2)/2
Sn=10(2+9(2))/2
Sn=10(2+18)/2
Sn=10(20)/2
Sn=200/2
Sn=100

Sum of the given numbers is 100.

hope it helps