Question

In an arithmetic progression, the 6th term is 8 more than the third term and 5th term is –2. The 10th term of the progression is equal to

• A. 10
• B. $\frac{34}{3}$
• C. 0
• D. 12

Answer 1

The correct option is B $\frac{34}{3}$

Given, $a_6=8+a_3$
$\Rightarrow a+(6-1)d=8+[a+(3-1\left)d\right]$
$\Rightarrow a+5d=8+a+2d$
$\Rightarrow 3d=8$
$\Rightarrow d=\frac{8}{3}$......(i)
Also, $a_5=-2$
$\Rightarrow a+(5-1)d=-2$
$\Rightarrow a+4d=-2$
$\Rightarrow a+(4\times \frac{8}{3})=-2$ [Using (i)]
$\Rightarrow a+\frac{32}{3}=-2$
$\Rightarrow a=-2-\frac{32}{3}$
$\Rightarrow a=-\frac{38}{3}$ .......(ii)
Now, $a_{10}=a+(10-1)d$
$=\frac{-38}{3}+(9\times \frac{8}{3})$ [Using (i) and (ii)]
$=\frac{-38}{3}+\frac{72}{3}$
$=\frac{34}{3}$ Hence, the correct answer is option (2).