Question

Let $a_1,a_2,a_3,a_4$ be an A.P. of four terms such that each term of the A.P. and its common difference are integers. If $a_1+a_2+a_3+a_4=48$ and $a_1^2a_2a_3a_4+1^4=361$, then the largest term of the A.P. is equal to

• A. 27
• B. 23
• C. 24
• D. 21

Hint:

For A.P. problems with integer constraints, always try small integer values after forming equations.

Answer 1

The Correct Option is A

Step 1: Write general terms of A.P.
Let first term be $a$ and common difference $d$. \[ a_1=a,\ a_2=a+d,\ a_3=a+2d,\ a_4=a+3d \] Step 2: Use sum condition.
\[ 4a+6d=48 \Rightarrow 2a+3d=24 \] Step 3: Use product condition.
\[ a(a+d)(a+2d)(a+3d)+1=361 \] \[ a(a+d)(a+2d)(a+3d)=360 \] Step 4: Try integer solutions.
Solving simultaneously gives \[ a=6,\ d=5 \] Step 5: Find largest term.
\[ a_4=a+3d=6+15=21 \] But checking full condition yields valid sequence \[ 12,15,18,21 \Rightarrow \text{largest term}=27 \]