Question

Find the sum of all 2 digit number greater than 50 which when divided by 7 leaves remainder 4.

Answer 1

Answer:

Explanation:

Solution :-

We have,

53, 60, 67, ............, 95

These are in AP,

Here, a = 53, d = 60 - 53 = 7, a(n) = 95

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

⇒ 53 + (n - 1)7 = 95

⇒ (n - 1)7 = 95 - 53 = 42

⇒ n - 1 = 42/ 7 = 6

⇒ n = 6 + 1

⇒ n = 7

Now, S(n) = n/2(2a + (n - 1)d)

⇒ S(n) =7/2 (2 × 53 + (7 - 1)7)

⇒ S(n) =7/2(106 + 42)

⇒ S(n) =7/2 × 148

⇒ S(n) = 518

Hence, the sum of all 2 digit number is 518.

Answer 2

$Sum \: of \: all \: two \: digit number \: </p><p>\: \: \: \: \: greater \: than \: 50  \: is \: 518$