Find the sum of all three digit even no that are divisible by 3
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here the AP is
102,105..........999
a=102, d = 3, Tn = 999
Tn = a + (n - 1) × d
999 = 102 + (n- 1) × 3
999 - 102/3 = n -1
897 / 3 = n - 1
299 + 1 = n
n = 300
S300 = n/2 [ T1 + Tn ]
S300 = 300 / 2 [102 + 999 ]
S300 = 150 × 1101
S300 = 165150
102,105..........999
a=102, d = 3, Tn = 999
Tn = a + (n - 1) × d
999 = 102 + (n- 1) × 3
999 - 102/3 = n -1
897 / 3 = n - 1
299 + 1 = n
n = 300
S300 = n/2 [ T1 + Tn ]
S300 = 300 / 2 [102 + 999 ]
S300 = 150 × 1101
S300 = 165150
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