Sum of all natural number between 300 and 500 divisible by 9
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Answer:
306,315,.......,495
A = 306
D = 9
Tn = a + (n - 1)d = 495
= 306 + 9n - 9 = 495
= 9n = 198
N = 22
S22 = n/2 (2a + (n - 1)d)
= 11 (612 + 189)
= 11 × 801
= 8811
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