what is the sum of all the two-digits numbers which when divided by 7 gives a remainder of 3?

According to question,
First Two Digit Number, which is divided by 7 and give the remainder 3, will be 10.
Last Two Digit Number, which is divided by 7 and give the remainder 3, will be 94.
The common difference between two consecutive numbers will be 7.
This series will be like →10, 17, 21 ,............................ 94.
Here First Number a = 10, Last number t_n = 94, Common Difference d = 7 ;
Using the Formula for Last Number t_n = a + (n - 1) x d;
⇒ a + (n - 1) x d = t_n
Put the value of a , d and t_n in above equation.
⇒ 10 + (n - 1) x 7 = 94
⇒ 10 + 7n - 7 = 94
⇒ 7n + 3 = 94
⇒ 7n = 94 - 3 = 91
⇒ n = 91/7
⇒ n = 13
Using the formula for the sum of Arithmetic Progression.
S_n = n/2 [ First Number + Last Number ] ;
Put the value of n , First Number and Last Number, we will get,
S_n = 13/2 [ 10 + 94 ]
S_n = 104 x 13/2
S_n = 52 x 13
S_n = 676