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Q : 10   Show that  \small a_1,a_2,...,a_n,...  form an AP where an is defined as below :

               (i)   \small a_n=3+4n

               Also find the sum of the first \small 15 terms .

Answers (1)

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It is given that 
\small a_n=3+4n
We will check values of a_n for different values of n
a_1 = 3+4(1) =3+4= 7
a_2 = 3+4(2) =3+8= 11
a_3 = 3+4(3) =3+12= 15
and so on.
From the above, we can clearly see that this is an AP with the first term(a) equals to and common difference (d) equals to 4
Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 2\times(7) +(15-1)4\right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 14 +56\right \}
\Rightarrow S_{15}= \frac{15}{2}\left \{ 70\right \}
\Rightarrow S_{15}= 15 \times 35
\Rightarrow S_{15}= 525
Therefore, the sum of 15 terms  is  525

Posted by

Gautam harsolia

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