find the sum of all 3 digit number which leave the remainder 2 when divided by 3
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This is a question of Arithmatic Sequence. et’s try and solve it in steps.
Step 1: Find the sequence
First term = 101; Last term =998Common difference = 3So, the sequence is:101, 104, 107, 110, ………, 998
Step 2: Find the number of terms
Hete [math]T_n = 998[/math]or, [math]101 + [(n - 1) \times 3] = 998[/math]or, [math]n = 299 + 1 = 300[/math]
Step 3: Find the Sum
[math]S_n = \frac {n}{2} \times [2a + (n - 1) \times d][/math][math]S_{300} = \frac {300}{2} \times [2 \times 101 + (300 - 1) \times 3][/math][math]S_{300} = 164850[/math] (Answer)
Step 1: Find the sequence
First term = 101; Last term =998Common difference = 3So, the sequence is:101, 104, 107, 110, ………, 998
Step 2: Find the number of terms
Hete [math]T_n = 998[/math]or, [math]101 + [(n - 1) \times 3] = 998[/math]or, [math]n = 299 + 1 = 300[/math]
Step 3: Find the Sum
[math]S_n = \frac {n}{2} \times [2a + (n - 1) \times d][/math][math]S_{300} = \frac {300}{2} \times [2 \times 101 + (300 - 1) \times 3][/math][math]S_{300} = 164850[/math] (Answer)
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