In a series, Tn=3-n then S5=?

1.	In a series, Tn=3-n then S5=?
Answer» $\mathfrak \red { \underline { \underline{hello \: friend \: here \: is \: your \: <klux>ANSWER</klux>}}} \\ a_{n} = 3 - n \\ so.. \\ a_{1} = 3 - 1 \\ a_{1} = 2 \\ a_{2} = 3 - 2 = 1 \\ so...d = - 1 \\ so.. s_{5} = \frac{5}{2} (2 \times 2 + (5 - 1)( - 1)) \\ s_{5} = \frac{5}{2} (4 - 4) \\ s_{5} = 0$ HOPE my answer will HELL you...

Answer»

$\mathfrak \red { \underline { \underline{hello \: friend \: here \: is \: your \: <klux>ANSWER</klux>}}} \\ a_{n} = 3 - n \\ so.. \\ a_{1} = 3 - 1 \\ a_{1} = 2 \\ a_{2} = 3 - 2 = 1 \\ so...d = - 1 \\ so.. s_{5} = \frac{5}{2} (2 \times 2 + (5 - 1)( - 1)) \\ s_{5} = \frac{5}{2} (4 - 4) \\ s_{5} = 0$

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In a series, Tn=3-n then S5=?​

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In a series, Tn=3-n then S5=?