✨Find the sum of first 123 even natural numbers.
✨Find the sum of all even numbers from 1 to 350.
Answers
Answer:
1.15252
2.30450
Explanation:
1. solution
The even natural numbers are 2, 4, 6, 8, … 27
The above sequence is an A.P. ∴ a = 2, d = 4 - 2 = 2, n = 123
Now, Sn = n/2 [ 2a + (n - 1)d]
∴ Sn = 123/2 [2(2) + (123 - 1)(2)] = 123/2 [2 (2) + 122 (2)]
= 123/2 x 2[2 + 122]
= 123 x 124 = 15252
∴ The sum of first 123 even natural numbers is 15252.
2. solution
The even numbers between 1 and 350 are 2, 4, 6,…, 348.
The above sequence is an A.P.
∴ a = 2, d = 4 – 2 = 2, tn = 348
Since, tn = a + (n – 1)d
∴ 348 = 2 + (n – 1)2
∴ 348 – 2 = (n – 1)2
∴ 346 = (n – 1)2
∴ n – 1 = 346/2
∴ n – 1 = 173
∴ n = 173 + 1 = 174
Now, Sn = n/2 [2a + (n – 1)d]
∴ S174 = 174/2 [2 (2) + (174 – 1)2] = 87(4 + 173 × 2)
= 87(4 + 346) = 87 × 350
∴ S174 = 30450
∴ The sum of all even numbers between 1 and 350 is 30450.
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Answer:
Find the sum of first 123 even natural numbers.
The even natural numbers are 2, 4, 6, 8, … 27
The above sequence is an A.P.
∴ a = 2, d = 4 - 2 = 2,
n = 123
Now, Sn = n/2 [ 2a + (n - 1)d]
∴ Sn = 123/2 [2(2) + (123 - 1)(2)]
= 123/2 [2 (2) + 122 (2)]
= 123/2 x 2[2 + 122]
= 123 x 124 = 15252
∴ The sum of first 123 even natural numbers is 15252
Find the sum of all even numbers from 1 to 350.
The even numbers between 1 and 350 are 2, 4, 6,…, 348.
The above sequence is an A.P.
∴ a = 2, d = 4 – 2 = 2, tn = 348
Since, tn = a + (n – 1)d
∴ 348 = 2 + (n – 1)2
∴ 348 – 2 = (n – 1)2
∴ 346 = (n – 1)2
∴ n – 1 = 346/2
∴ n – 1 = 173
∴ n = 173 + 1 = 174
Now, Sn = n/2 [2a + (n – 1)d]
∴ S174 = 174/2 [2 (2) + (174 – 1)2]
= 87(4 + 173 × 2)
= 87(4 + 346) = 87 × 350
∴ S174 = 30450
∴ The sum of all even numbers between 1 and 350 is 30450.