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(A) Let the four terms be $a-3d, a-d, a+d, a+3d$ where the common difference is $l=2d$.Since the sum is $48$, we have $(a-3d)+(a-d)+(a+d)+(a+3d)=48$,

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(A) Let the four terms be $a-3d, a-d, a+d, a+3d$ where the common difference is $l=2d$.Since the sum is $48$, we have $(a-3d)+(a-d)+(a+d)+(a+3d)=48$, which simplifies to $4a=48$, so $a=12$.The product of the terms plus $l^4$ is given by $(a^2-9d^2)(a^2-d^2)+l^4=361$.Since $l=2d$, $l^4=16d^4$. Substituting $a=12$:$(144-9d^2)(144-d^2)+16d^4=361$$20736 - 144d^2 - 1296d^2 + 9d^4 + 16d^4 = 361$$25d^4 - 1440d^2 + 20375 = 0$Dividing by $5$: $5d^4 - 288d^2 + 4075 = 0$.Using the quadratic formula for $d^2$: $d^2 = \frac{288 \pm \sqrt{288^2 - 4(5)(4075)}}{10} = \frac{288 \pm \sqrt{82944 - 81500}}{10} = \frac{288 \pm \sqrt{1444}}{10} = \frac{288 \pm 38}{10}$.$d^2 = \frac{326}{10} = 32.6$ (not an integer $d$) or $d^2 = \frac{250}{10} = 25$.Thus $d=5$ (since $d$ must be an integer for $l=2d$ to be an integer).The terms are $12-15, 12-5, 12+5, 12+15$, which are $-3, 7, 17, 27$.The largest term is $27$.

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