Question
How do I solve 1 + 3 + 5 + 7 +… + 101?
A. 2421
B. 2601
C. 2341
D. 2701
Answer: B. 2601
Solution:
This series is an arithmetic progression
1, 3, 5, 7, ………, 101
Having First No (a) = 1, Last No (tn) = 101 and
Common difference (d) = Previous no – Next No = 3 – 1 = 5 – 3 = 7 – 5 = 2
According to formula
tn = [a + (n-1) d] where n is no of terms
=> 101 = 1 + (n – 1) * 2
=> 101 – 1 = (n – 1) * 2
=> 100 / 2 = n – 1
=> 50 + 1 = n
=> n = 51
According to sum formula
Sn = [n * (a + l) ] / 2
= [51 * (1 + 101) ] / 2
= (51 * 102) / 2
= 51 * 51
Sn = 2601
Hence, The answer is 2601.
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