Question

find the sum of all multiples of 7 laying between 100 and 1000​

Answer 1

Answer:

70336

Step-by-step explanation:

between 100 and 1000

d= 7 as the numbers are multiples of 7

the first number divisible by 7 = a =105

the last number divisible by 7 = l = 994

l = a+(n-1)d

994= 105 + (n-1)7

994-105= (n-1)7

889/7= n-1

127+1=n

n=128

128 numbers between 100 and 1000 are divisible by 7

sum= (n/2)(a+l)

= (128/2)(105+994)

= 64×1099

= 70336

Answer 2

Multiples of 7 b/w 100 and 1000 are:

105, 112, 119, 126, ........, 994.

Clearly, the above listed series is an AP with:

First term (a) = 105

Common Difference (d) = 112 - 105 = 7

Last Term (b) = 994

Now, all we need to do is- find the sum of the given AP. And our objective can be achieved by applying the relation:

Sum (Sn) = n/2(a + b)

But the problem arises that- what about 'n'. It's unknown. So, Let's try and calculate it.

We know that, last term (b) = 994

=> a + (n-1)d = 994

105 + (n-1)7 = 994

(n-1)7 = 889

(n-1) = 127

n = 128

Therefore the Sum, S(n) = n/2(a + b)

=> S = 128/2(105 + 994)

= 64(1099)

= 70336

Hence, the required sum is 70336.