Question
Find the sum:
$3 + 11 + 19 + ..... + 803.$
$3 + 11 + 19 + ..... + 803.$
✓
Answer
$3 + 11 + 19 + ..... + 803$
$a = 3, d = 11 - 3 = 8, l = a_n = 803$
$803 = 3 + (n - 1)8$
$\frac{800}{8}=\text{n}-1$
$\text{n}=101$
$\text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})$
$\text{S}_{101}=\frac{101}{2}(3+803)$
$\text{S}_{101}=\frac{101}{2}(806)$
$\text{S}_{101}=101\times403$
$=40703$
$a = 3, d = 11 - 3 = 8, l = a_n = 803$
$803 = 3 + (n - 1)8$
$\frac{800}{8}=\text{n}-1$
$\text{n}=101$
$\text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})$
$\text{S}_{101}=\frac{101}{2}(3+803)$
$\text{S}_{101}=\frac{101}{2}(806)$
$\text{S}_{101}=101\times403$
$=40703$
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