The sum of 1+3+5+7+...+101=?​

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

B-simple yes/no questions