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  4. If a1 , a2 , a3 ,..... are in A.P. such that a1 + a7 + a16 = 40, then the sum of the first 15 terms of this A.P. is :

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Q. If a$_1 , a_2 , a_3$ ,..... are in $A.P$. such that a$_1 + a_7 + a_{16} = 40$, then the sum of the first $15$ terms of this $A.P$. is :

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Solution:

$a_1 + a_7 + a_{16} \, = 40$
$a + a + 6d + a + 15d = 40$
$\Rightarrow \, \, 3a + 21d = 40 \, \, \, \, \, \, \, \Rightarrow a + 7d = \frac{40}{3}$
$S_{15} \, = \, \frac{15}{2} (2a + 14d) \, = 15 (a+7d)$
$ \, \, \, \, \, S_{15} = 15 \times \frac{40}{3} \Rightarrow \, \, 200 \, \, \, S_{15}= 200$