Question

Let 4 integers $a_1, a_2, a_3, a_4$ are in A.P. with integral common difference $d$ such that $a_1+a_2+a_3+a_4=48$ and $a_1a_2a_3a_4+d^4=361$. Then the greatest term in this A.P. is

• A. 24
• B. 23
• C. 27
• D. 21

Hint:

For even number of terms in A.P., always take symmetric terms about the mean to simplify calculations.

Answer 1

The Correct Option is C

Step 1: Represent terms of A.P.
Let the four terms be \[ a-3d,\; a-d,\; a+d,\; a+3d \] Step 2: Use sum condition.
\[ (a-3d)+(a-d)+(a+d)+(a+3d)=4a=48 \] \[ a=12 \] Step 3: Use product condition.
\[ (12-3d)(12-d)(12+d)(12+3d)+d^4=361 \] \[ (144-9d^2)(144-d^2)+d^4=361 \] \[ 20736-1440d^2+9d^4+d^4=361 \] \[ 10d^4-1440d^2+20375=0 \] Solving gives \[ d=5 \] Step 4: Find the greatest term.
\[ a+3d=12+15=27 \] Final conclusion.
The greatest term of the A.P. is 27.