Question

find the sum of all natural numbers between 100 and 800 which are divisible by 7​

Answer 1

The first number after 100 which is divisible by 7 is 105 and the last term is 798.

Here a= 105 and ; l= 798

l= a+ (n-1)d

798=105 + (n-1)*7

n=100

we know that sum of arithmetic progression is s= $\frac{n}{2} (a+l)$

   Implies s=50*903

                 s=4515

Answer 2

⠀⠀⠀⠀ ⠀✁ To find...

⠀⠀The sum of natural numbers between 100 amd 800 which are divisible by 7...

⠀⠀⠀⠀ ⠀✁ SoLution...

⠀A.P. :- 105, 112, 119, 126.........798

• an = a + ( n - 1 ) × d

---✁ 798 = 105 + ( n - 1 ) × 7

---✁ 798 = 105 + 7n - 7

---✁ 798 = 98 + 7n

---✁ 798 - 98 = 7n

---✁ 700 = 7n

---✁ 100 = n

Total numbers between 100 and 700 which divisible by 7 are 100...

• Sn = n / 2 ( a + l )

---✁ S100 = 100 / 2 ( 105 + 798 )

---✁ S100 = 50 × 903

---✁ S100 = 45,150

The sum of all numbers between 100 and 800 is 45,150...

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