The natural number divisible by 4 between 1 to 140 is

4, 8, 12, …….136

Where first term a = 4

Second term t_{1} = 8

Third term t_{2} = 12

Thus, common difference d = t_{2} – t_{1} = 12 – 8 = 4

t_{n} = 136

Now, by using n^{th} term of an A.P. formula

t_{n} = a + (n – 1) d

where n = no. of terms

a = first term

d = common difference

t_{n} = n^{th} terms

we can find value of “n” by substituting all the value in formula we get,

⇒ 136 = 4 + (n – 1) × 4

⇒ 136 – 4 = 4(n – 1)

⇒ 132 = 4(n – 1)

⇒ n – 1 = 132/4 = 33

⇒ n = 33 + 1 = 34

Now, by using sum of n^{th} term of an A.P. we will find its sum

Where, n = no. of terms

a = first term

d = common difference

S_{n} = sum of n terms

Thus, substituting given value in formula we can find the value of S_{34}

⇒S_{34} = 17 × [8 + 33×4]

⇒S_{34} = 17 × [8 + 132]

⇒S_{34} = 17 × 140 = 2380

Thus, S_{34} = 2380