Question

find the sum of all the three digit natural number which are divisible of 7.

Answer 1

Heya mate, Here is ur answer

a(first term) = 105

d =7 (divisiblity by 7)

an= 994

a+(n-1)d=994

105+(n-1)7=994

(n-1) 7=994-105

(n-1) 7=889

n-1=889/7

n-1=127

n=127+1

n=128

Sn=n/2 (a+an)

$= \frac{128}{2} (105 + 994)$


$= 64 \times 1099$


$= 70336$


$<b><u>So, sum of all 3 digits number divisible by 7=70336 </u></b>$

$<b><u><i>❤Hope it helps uh ❤</b></u></i>$

==========================

Warm regards

@Laughterqueen

Be Brainly ✌✌✌

Answer 2

Answer: 70336

Step-by-step explanation:

Let the A.P be

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

a = 105 ; d = 112-105 = 7 ; l = 994

a + (n - 1)d = 994

105 + (n - 1)7 = 994

105 + 7n - 7 = 994

7n = 994 - 105 + 7

7n = 889 - 7

7n = 889

n = 128

Now,

a = 105 ; d = 7 ; n = 128

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

Sn = 128/2[2×105 + (128 - 1)7]

Sn = 64[210 + 896 - 7]

Sn = 64[1106 - 7]

Sn = 64[1099]

Sn = 70336