QUESTION 5
Easy
How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
SOLUTION
1
Write the AP
Two-digit multiples of 3: 12, 15, 18, ..., 99
Here,
first term, \(a = 12\)
common difference, \(d = 3\)
last term, \(l = 99\)
Here,
first term, \(a = 12\)
common difference, \(d = 3\)
last term, \(l = 99\)
2
Find the number of terms
\(l = a + (n − 1)d\)
⇒ \(99 = 12 + (n − 1)3\)
⇒ \(87 = 3(n − 1)\)
⇒ \(29 = n − 1\)
⇒ \(n = 30\)
So there are 30 terms in the AP.
⇒ \(99 = 12 + (n − 1)3\)
⇒ \(87 = 3(n − 1)\)
⇒ \(29 = n − 1\)
⇒ \(n = 30\)
So there are 30 terms in the AP.
3
Find the sum
\(S_n = \frac{n}{2}(a + l)\)
⇒ \(S_{30} = \frac{30}{2}(12 + 99)\)
⇒ \(S_{30} = 15(111)\)
⇒ \(S_{30} = 1665\)
The sum is 1665.
⇒ \(S_{30} = \frac{30}{2}(12 + 99)\)
⇒ \(S_{30} = 15(111)\)
⇒ \(S_{30} = 1665\)
The sum is 1665.
🏆
Final Answer : The sum of all the two-digit multiples of 3 is 1665.
Concept Note
The sum of an AP is given by the formula
\(S_n = \frac{n}{2}(a + l)\)
or
\(S_n = \frac{n}{2}[2a + (n − 1)d]\)
where
\(n = \)number of terms
\(a = \)first term
\(l = \)last term
\(d = \)common difference