Question

please answer this question ​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

multiples of 5 are 5,10,15,20,25.........so on

multiples of 5 btw 100 and 1000 are

105,110,115,120,..........990,995

Since the diff btw the consecutive terms is constant,this belongs to AP(arithmetic progression)

so,first term,a=105

last term,l=5

common difference,d=110-105=5

Lets first find no. of terms,n

an=a+(n-1)d an is nth term which is obviously the last term here=995 i.e. l

So lets just quickly put the values..

995=105+(n-1)5

995=105+5n-5

990=105+5n

895=5n

n=179

To find sum ,use the formula

Sn=n[2a+(n-1)d]/2

here n=179,a=105,d=5

=179[2(105)+(179-1)5]/2

=179(210+178×5)/2

=179(1100)/2

=98450 ans

so ur ans is 98450=sum