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Q : 3      In an AP:

              (iii) given  \small a_1_2=37,d=3, find \small a and \small S_1_2.
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It is given that
d = 3 \ and \ a_{12} = 37
a_{12}= a+11d = 37
        = a= 37-11\times 3 = 37-33=4
Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{12} = \frac{12}{2}\left \{ 2\times(4) +(12-1)3\right \}
\Rightarrow S_{12} = 6\left \{ 8+33\right \}
\Rightarrow S_{12} = 6\left \{41\right \}
\Rightarrow S_{12} =246
Therefore, the sum of given AP  is 246

Posted by

Gautam harsolia

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