Question

If sum of the $3^{rd}$ and the $8^{th}$ terms of an AP is 7 and the sum of the $7^{th}$ and the ${14}^{th}$ terms is 3, find the ${10}^{th}$ term to closest integer.

Answer 1

Let $a$ and $d$ be the first term and common difference of AP respectively & nth term is given as $a+(n-1)d$
Since $a_3+a_8=7\Rightarrow (a+2d)+(a+7d)=2a+9d=7$ ...(1)
$a_7+a_{14}=3\Rightarrow (a+6d)+(a+13d)=2a+19d=3$ ...(2)
Subtracting (1)from(2),we get
$10d=-4$
$\Rightarrow d=\frac{-2}{5}$
putting $d=\frac{-2}{5}$ in (1),we get
$2a+9\times \frac{-2}{5}=7$
$\Rightarrow 2a=\frac{53}{5}$
$\Rightarrow a=\frac{53}{10}$
Now, $a_{10}=a+9d=\frac{53}{10}+9\times \frac{-2}{5}$
$\Rightarrow a_{10}=\frac{17}{10}=1.7=2$