The sum of four numbers in an arithmetic progression is 48 and the ratio of the product of the extremes to the product of the two middle terms is…

Question

The sum of four numbers in an arithmetic progression is 48 and the ratio of the product of the extremes to the product of the two middle terms is 27:35, then the largest term of this arithmetic progression is

MARKS Content Team · Accepted Answer

Correct answer: (D) 18. Let the A. P. be a-3d, a-d, a+d, a+3d The sum of the terms =48=4a⇒a=12 Given, (12-3d12+3d)/(12-d12+d)=27/35 ⇒(94-d4+d)/(12-d12+d)=27/35 ⇒16-d^235=144-d^23 ⇒35d^2-3d^2=16×35-144×3 ⇒32d^2=1635-27=16×8 ⇒d^2=4⇒d=±2 So the numbers are: 6,10,14,18

The sum of four numbers in an arithmetic progression is $48$ and the ratio of the product of the extremes to the product of the two middle terms is $27 : 35$ , then the largest term of this arithmetic progression is

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