The sum of four terms in A.P is $36$ .If the product of the second and the third terms is $8$ more that the product of the fourth terms, find the four terms ?

5, 8, 10, 12

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7, 8, 10, 11

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6, 8, 10, 12

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4, 6, 10, 12

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Solution

The correct option is C $6, 8, 10, 12$
$S_{4} = 36 \Rightarrow \frac{4}{2} [2 a + (4 - 1) d] = 36 \Rightarrow 2 a + 3 d = 18 ⟶ (1) A / Q, a_{2} . a_{3} = a_{1} . a_{4} + 8 \Rightarrow (a + d) (a + 2 d) = a . (a + 3 d) + 8 \Rightarrow d^{2} = 4 \Rightarrow d = \pm 2$
If $d = 2$ then $a = 6$
then the four terms is $6, 8, 10, 12$
If $d = - 2$ then $a = 12$
then the four terms is $12, 10, 8, 6$

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The sum of four terms in A.P is 36.If the product of the second and the third terms is 8 more that the product of the fourth terms, find the four terms ?

The correct option is C 6,8,10,12S4=36⇒42[2a+(4−1)d]=36⇒2a+3d=18⟶(1)A/Q,a2.a3=a1.a4+8⇒(a+d)(a+2d)=a.(a+3d)+8⇒d2=4⇒d=±2If d=2 then a=6then the four terms is 6,8,10,12If d=−2 then a=12then the four terms is 12,10,8,6

The sum of four terms in A.P is $36$ .If the product of the second and the third terms is $8$ more that the product of the fourth terms, find the four terms ?