The two digit natural numbers which leave a remainder 5, when divided by 7 are 12, 19, 26 ...., 89, 96.
∴ 12, 19, 26, ...., 89, 96 is an A.P. whose first term a = 12 and common difference d = 7.
Let the last or nth term be Tn.
Then, Tn = a + (n – 1) d, where n is the number of terms in A.P.
⇒ 96 = 12 + (n – 1) 7
⇒ 84 = (n – 1) 7 ⇒ n – 1 = 12 ⇒ n = 13
∴ Required Sum = \(\frac{n}{2}\) (a + l) = \(\frac{13}{2}\)(12 + 96)
= \(\frac{13}{2}\) × 108
=13 × 54 = 702.
