Find the number of perfect squares in the given series 2013, 2020, 2027,................, 2300 (Hint 44^2=1936) a. 1 b. 2 c. 3 d. Can’t be determined Answer: a Explanation: The given series is an AP with common difference of 7. So the terms in the above series are in the form of 2013 + 7k. We have to find the perfect squares in this format in the given series. Given that 44^2 = 1936. Shortcut: To find the next perfect square, add 45th odd number to 44^2. So 45^2 = 1936 + (2 x 45 -1) = 2025 46^2 = 2025 + (2 x 46 - 1) = 2116 47^2 = 2116 + (2 x 47 - 1) = 2209 Now subtract 2013 from the above numbers and divide by 7. Only 2209 is in the format of 2013 + 7k. One number satisfies.
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