Question
The sum of all even numbers between 21 and 51 is
Answer: Option C
Answer: (c)22 + 24 + 26 + ... + 50 = 2 (11 + 12 + 13 + .... + 25) = 2 [(1 + 2 + 3 + ... + 25) – (1 + 2 + 3 ... + 10)] = 2$({25×26}/2 - {10×11}/2)$= 2 (325 – 55) = 2 × 270 = 540 Method 2 :Sum of first n even numbers = n (n + 1) Required sum = Sum of 25 even numbers from 1 to 50 – sum of 10 even numbers from 1 to 20 = 25×26 – 10 × 11 = 650 – 110 = 540
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Answer: (c)22 + 24 + 26 + ... + 50 = 2 (11 + 12 + 13 + .... + 25) = 2 [(1 + 2 + 3 + ... + 25) – (1 + 2 + 3 ... + 10)] = 2$({25×26}/2 - {10×11}/2)$= 2 (325 – 55) = 2 × 270 = 540 Method 2 :Sum of first n even numbers = n (n + 1) Required sum = Sum of 25 even numbers from 1 to 50 – sum of 10 even numbers from 1 to 20 = 25×26 – 10 × 11 = 650 – 110 = 540
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