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The sum of the third and seventh terms of an AP is $40$ and the sum of its sixth and fourteenth terms is $70$. Find the sum of the first ten terms of the AP.
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$a_3 + a_7 = 40$
$(a + 2d) + (a + 6d) = 40$
$\Rightarrow 2a + 8d = 40$ or $a + 4d = 20$ --- (1)
$a_6 + a_{14} = 70$
$(a + 5d) + (a + 13d) = 70$
$\Rightarrow 2a +18d = 70$ or $a + 9d = 35$ --- (2)
Solving (1) and (2), we get
$a = 8$ and $d = 3$
$S_{10} = \frac{10}{2} \times [2 (8) + 9 (3)]$
$= 215$
$(a + 2d) + (a + 6d) = 40$
$\Rightarrow 2a + 8d = 40$ or $a + 4d = 20$ --- (1)
$a_6 + a_{14} = 70$
$(a + 5d) + (a + 13d) = 70$
$\Rightarrow 2a +18d = 70$ or $a + 9d = 35$ --- (2)
Solving (1) and (2), we get
$a = 8$ and $d = 3$
$S_{10} = \frac{10}{2} \times [2 (8) + 9 (3)]$
$= 215$