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Question
Find out the sum of all natural numbers between 1 and 145 which are divisible by 4.
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Solution
The numbers divisible by 4 between 1 and 145 are
4, 8, 12, 16, .........144 ; which is an A. P.
Here, a = 4, d = 4, tn = 144 we have to find n.
tn = a + (n - 1) d
∴tn = 4 + (n - 1) × 4
∴ 144 = 4n
∴ n = 36
Now, `s_n = n/2[t_1+t_n]`
`∴ S_36 = 36/2 [4+144]`
= 18 × 148 = 2664
Alternate Method
4 + 8 + 12 + ..... + 144
= 4(1 + 2 + 3 + ..... + 36)
`= (4xx36xx37)/2`
= 12 × 6 × 37
= 444 × 6
= 2664
This is also possible.
∴ The sum of numbers between 1 and 145 divisible by 4 is 2664.
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