Question

Find the sum of all even numbers between 1 and 350.

Answer 1

The given sequence is 2, 4, 6, 8 ... 344, 346, 348

Now, surely the above-mentioned range is in an A.P. or Arithmetic Progression. (Because the difference between two consecutive numbers in the given series is always constant.)

The above sequence is an A.P. with

a = t₁ = 2,

d = t₂ - t₁

= 4 - 2

= 2

t_n = 348

Since t_n = a + (n - 1) d

∴ 348 = 2 + (n - 1) 2

∴ 348 = 2 - 2n + 2

∴ 348 = 2 + 2n - 2

∴ 348 = 2n

∴ n = `348/2`

∴ n = 174

Thus, the number of terms (n) = 174.

Now, S_n = `n/2` [2a + (n - 1)d]

∴ S₁₇₄ = `174/2` [2(2) + (174 - 1)2]

= 87 [4 + (173)2]

= 87 [4 + 346]

= 87 × 350

= 30450

Hence, the sum of all even numbers between 1 and 350 is 30450.

Answer 2

All even numbers between 1 and 350,

2, 4, 6, 8 ... 344, 346, 348

Given sequence is an A.P.

a = 2, d = 2, t_n = 348

t_n = a + (n - 1)d

348 = 2 + (n - 1)2

348 - 2 = (n - 1)2

`346/2` = n - 1

173 + 1 = n 

n = 174

t₁ = 2, t_n = 348

S_n = `n/2[t_1+t_n]`

S₁₇₄ = `174/2[2+348]`

S₁₇₄ = 87 × 350

S₁₇₄ = 30450

Hence, the sum of all even numbers between 1 and 350 is 30450.