Find the sum of 1+3+5+....+45


A) 540

B) 539

C) 529

D) 579

Correct Answer:
C) 529

Description for Correct answer:
1+3+5+....+45

Here a=1, d=2, l=45

\( \Large n=\frac{l-a}{d}+1 = \frac{45-1}{2}+1 = 22+1 = 23\)

Therefore, 1+3+5+....+45 = \( \Large n^{2} \)

=\( \Large 23^{2} = 529 \)

Part of solved Factorisation questions and answers : >> Elementary Mathematics >> Factorisation








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