Q : 3     In an AP:

            (ii) given \small a=7\small a_1_3=35, find \small d and \small S_1_3

Answers (1)

It is given that
a = 7 \ and \ a_{13} = 35
a_{13}= a+12d = 35
        = 12d = 35-7 = 28
d = \frac{28}{12}= \frac{7}{3}
Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{13} = \frac{13}{2}\left \{ 2\times(7) +(13-1)(\frac{7}{3})\right \}
\Rightarrow S_{13} = \frac{13}{2}\left \{14 +28\right \}
\Rightarrow S_{13} = \frac{13}{2}\left \{42\right \}
\Rightarrow S_{13} = 13 \times 21 = 273
Therefore, the sum of given AP  is 273

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