Question

Find the sum of numbers between 1 to 140, divisible by 4

Answer 1

The numbers between 1 to 140 divisible by 4 are

4, 8, 12, ......, 140

The above sequence is an A.P.

∴ a = 4, d = 8 - 4 = 4

Let the number of terms in the A.P. be n.

Then, t_n = 140

Since t_n = a + (n – 1)d,

140 = 4 + (n – 1)(4)

∴ 140 - 4 =  (n – 1)(4)

∴ 136 = (n – 1)(4)

∴ ${\displaystyle \frac{136}{4}}$  = n - 1

∴ 34 + 1 = n

∴ n = 35

Now, S_n = ${\displaystyle \frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}$

∴ S₃₅ = ${\displaystyle \frac{35}{2}[2\times 4+(35-1)4]}$

= ${\displaystyle \frac{35}{2}[8+\left(34\right)4]}$

= ${\displaystyle \frac{35}{2}[8+136]}$

= ${\displaystyle \frac{35}{2}\times 144}$

= 35 × 72

S₃₅ = 2520

∴ The sum of numbers between 1 to 140, which are divisible by 4 is 2520.