Plz its urgent solve this find the sum of two digit number divisible by 2 or 3.plz fsst
Answers
Answer:
3285
Step-by-step explanation:
(1)
Two digit numbers divisible by 2:
Series is 10,12,14...98
First term is 10.
Common difference is 2.
Last term an = 98.
nth term of an AP an = a + (n - 1) * d
98 = 10 + (n - 1) * 2
98 = 10 + 2n - 2
98 = 8 - 2n
90 = 2n
n = 45
Sum of first n terms = n/2[2a + (n - 1) * d]
= 45/2[20 + 44 * 2]
= 45/2[108]
= 2430.
(2)
Two digit numbers divisible by 3:
12,15,18... 99
First term a = 12
Common difference d = 3
Last term an = 99.
nth term of AP = a + (n - 1) * d
99 = 12 + (n - 1) * 3
99 - 12 = 3n - 3
90 = 3n
n = 30
Sum of first n terms = n/2[2a + (n - 1) * d]
= 30/2[24 + (30 - 1) * 3]
= 1665
(3)
Two digit numbers divisible by 6:
12,18,24,..96
First term a = 12
Common difference d = 6
Last term an = 96.
nth term of an AP an = a + (n - 1) * d
96 = 12 + (n - 1) * 6
96 - 12 = 6n - 6
90 = 6n
n = 15
Sum of first n terms = n/2[2a + (n - 1) * d]
= 15/2[24 + 84]
= 810.
So, Sum of two digit numbers = 2430 + 1665 - 810
= 3285
Hence, Sum of two digit numbers divisible by 2 or 3 = 3285.
Hope it helps you
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