What is the sum of all two-digit numbers which leave remainder 5 when

1.	What is the sum of all two-digit numbers which leave remainder 5 when they are divided by 7?
Answer» 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 T_n. Then, T_n = 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.

What is the sum of all two-digit numbers which leave remainder 5 when they are divided by 7?

Answer»

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 T_n.

Then, T_n = 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.