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An arithmetic progression is written in the following way [figure] The sum of all the terms of the 10th row is .
1505
The terms are the AP 2, 5, 8, 11, (d = 3); row n has n terms
Before row 10: 1 + 2 + + 9 = 45 terms
first term of row 10 = T46 = 2 + 45 3 = 137
Row 10: 137, 140, , 137 + 9 3 = 164 (10 terms)
∴ Sum = 102(137 + 164) = 5 301 = 1505
Answer: 1505
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