Question

The four numbers in A.P such that the sum of the first and last number is $8$ and the product of second and third number is $12$ are

• A. $3,6,2,5$
• B. $2,3,4,6$
• C. $-2,2,6,10$
• D. None of these

Answer 1

The correct option is C $-2,2,6,10$

The four numbers in A.P be $a-3d,a-d,a+d,a+3d$

Given $(a-3d)+(a+3d)=8$

$⟹2a=8⟹a=4$

Now, $(a-d).(a+d)=12$

$⟹(4-d)(4+d)=12$

$⟹16-d^2=12$

$⟹d^2=4⟹d=2$

Hence numbers are $-2,2,6,10$.