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Given :   S_{n}=615, n=41, a_{n}=22, a_{1}=?

First, lets find the first term of this sequence using the arithmetic series.
               S_{n}= \frac{n}{2} (a_{1}+a_{n})
               615= \frac{41}{2} (a_{1}+22)
               615= \frac{41}{2}a_{1}+ \frac{902}{2}
               615= \frac{41}{2}a_{1} + 451
                \frac{164}{ \frac{41}{2} } = a_{1}
               164( \frac{2}{41})=a_{1}

since we find already the first term, we will find the common difference.

                \frac{14}{40} or \frac{7}{20}=d
              a_{2}=8+ \frac{7}{20}
              a_{2}=8 \frac{7}{20}

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