#### Question

Find out the sum of all natural numbers between 1 and 145 which are divisible by 4.

#### 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.

t_{n} = a + (n - 1) d

∴t_{n} = 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.

Is there an error in this question or solution?

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Find Out the Sum of All Natural Numbers Between 1 and 145 Which Are Divisible by 4. Concept: Sum of First n Terms of an AP.

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