Question

find 5+10+15+...............+1000​

Answer 1

Answer:

The sum of 5+10+15 ….+100 is the sum of an AP whose first terms is 5, the last term is 100 and the common difference is 5.

Th = 100 = a +(n-1)d = 5 +(n-1)5 = 5

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

= (20/2)[2*5+ (20–1)*5]

= 10[10+95]

= 10 × 105

= 1050

Step-by-step explanation:

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Answer 2

Answer:

Here it is

a1 = 5

a2 = 10

d = a2 - a1

d= 10 - 5

d= 5