Find the 11 and12 numbered question.
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Q11) Natural numbers between 1 and 100 which are divisible by 3 are 3, 6, 9, 12, ……….99
Here, first term , a = 3, last term, l = 99, common Difference, d = 6 - 3 = 3
By using the formula ,an = a + (n - 1)d
99 = 3 + (n - 1)3
99 = 3 + 3n - 3
99 = 3n
n = 99/3
n = 33
By using the formula ,Sum of nth terms , Sn = n/2 [a + l]
S33 = 33/2 [3 + 99]
S33 = 33/2 × 102
S33 = 33 (51)
S33 = 1683
Hence, the sum of all natural numbers between 1 and 100, which are divisible by 3 are 1683.
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