Find the sum of 1+3+5+....+45
Correct Answer: 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 = \( \Large n^{2} \)
=\( \Large 23^{2} = 529 \)
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