Question

In an A.P , the fourth term exceeds four times the $12th$ term by one and the third term exceeds twice the tenth term by five, find the A.P.

• A. $27,25,23,21,19,17,15,\mathrm{13.........}$
  
• B. $1,3,5,7,11,13,17,19,......31$
• C. Data insufficient
  
• D. None of these
  

Answer 1

The correct option is A $27,25,23,21,19,17,15,\mathrm{13.........}$

According to the question
$a_4=4a_{12}+1$
$a_1+3d=4(a_1+11d)+1$
$a_1+3d=4a_1+44d+1$
$-3a_1=41d+1$
$-a_1=\frac{41d+1}{3}$ ............(1)
$a_3=2a_{10}+5$
$a_1+2d=2(a_1+9d)+5$
$a_1+2d=2a_1+18d+5$
$-a_1=16d+5$
From eq (1)
$\frac{41d+1}{3}=16d+5$
$41d+1=48d+15$
$-7d=14$
$d=-2$
$a_1=\frac{41d+1}{-3}=\frac{41\times -2+1}{-3}=21$
$\therefore$ series in AP $=27,25,23,21$