To find the sum of an arithmetic series, we use the formula:
\[ \text{Sum} = \frac{n}{2}(a_1 + a_n) \]
where \(n\) is the number of terms in the series, \(a_1\) is the first term, and \(a_n\) is the last term.
In this case, we have \(n = 140\), \(a_1 = 1\), and \(a_n = 140\). Substituting these values into the formula, we get:
\[ \text{Sum} = \frac{140}{2}(1 + 140) = 70 \times 141 = 9870 \]
To find the greatest prime divisor of 9870, we can factorize it:
\[ 9870 = 2 \times 3 \times 5 \times 7 \times 47 \]
The greatest prime divisor of 9870 is 47.
Therefore, the greatest prime divisor of the arithmetic series 1 + 2 + 3 + ... + 135 + 136 + 137 + 138 + 139 + 140 is \(\boxed{47}\).
Find the greatest prime divisor of the value of the arithmetic series
1 + 2 + 3 + \dots + 135 + 136 + 137 + 138 + 139 + 140.
1 answer