Question

What is the sum of all two-digit numbers which when divided by 3 leave 1 as the remainder?
1. 1234
2. 1605
3. 1650
4. None of these

Answer 1

Correct Answer - Option 2 : 1605

Concept:

The formula for finding the nth term of an AP is:

an = a + (n − 1) \(\rm \times \) d

Where 

a = First term

d = Common difference

n = number of terms

an = nth term

 

The formula for finding the sum of n Terms of an AP is:

S = \(\rm \frac {n}{2}\) [2a + (n − 1) \(\rm \times \) d]

S  = \(\rm \frac {n}{2}\) (first term + last term)

 

Calculation:

Clearly, the 2-digit number which when divided by 3 gives remainder 1 are: 10, 13, 16, ... 97
Here, a = 10, d = 13 - 10 = 3, l = 97
t_n = 97
n^th term of an AP is t_n = a + (n – 1)\(\rm \times \)d
97 = 10 + (n – 1)\(\rm \times \)3

97 = 10 + 3n – 3

97 = 7 + 3n

3n = 97 – 7

3n = 90

n = 90/3

n = 30

Sum of n terms of AP, 

S_n = \(\rm \frac{n}{2}\)(a + l)

= \(\rm \frac{30}{2}\)\(\rm \times \) (10 + 97)

= 15 \(\rm \times \) 107

= 1605
Hence sum of 2-digit numbers which when divided by 3 yields 1 as the remainder is 1605.

 

These formulas are useful to solve problems based on the series and sequence concept.

General Form of AP	a, a + d, a + 2d, a + 3d, . . .
The nth term of AP	an = a + (n – 1) \(\rm \times \) d
Sum of n terms in AP	S = \(\rm \frac{n}{2}\)[2a + (n − 1) \(\rm \times \) d]
Sum of all terms in a finite AP with the last term as ‘l’	\(\rm \frac{n}{2}\)(a + 1)