Question

The sum of the $4$th and $8$th terms of an AP is $24$ and the sum of the $6$th and $10$th terms is $44.$ Find the first three terms of the AP.

• A. The first three terms of the AP are $t_1=18,t_2=12,t_3=6$
  
• B. The first three terms of the AP are $t_1=-12,t_2=-2,t_3=8$
  
• C. The first three terms of the AP are $t_1=-13,t_2=-8,t_3=-3$
• D. None of these
  

Answer 1

The correct option is C The first three terms of the AP are $t_1=-13,t_2=-8,t_3=-3$

Let the first term of an A.P $=a$
And the common difference of the given A.P $=d$
$a_n=a+(n-1)d$
$a_4=a+(4-1)d=a+3d$
Similarly
$a_8=a+7d$
$a_6=a+5d$
$a_{10}=a+9d$
Sum of 4th and 8th term $=24$
$\therefore a_4+a_8=24$
$a+3d+a+7d=24$
$2a+10d=24$
$a+5d=\mathrm{12...1}$
Sum of 6th and 10th term $=44$
$\therefore a_6+a_{10}=44$
$a+5d+a+9d=44$
$2a+14d=44$
$a+7d=\mathrm{22...}ii$
Solving i and ii we get
$a+7d-a-5d=22-12$
$2d=10$
$d=5$
From eq (i) we get
$a+5d=12$
$a+5\times 5=12$
$\therefore a=-13$
$\Rightarrow a_2=a+d=-13+5=-8$
$\Rightarrow a_3=a_2+d=-8+5=-3$
$\therefore$ The fires terms of this A.P are $-13,-8,-3$