Question

(ii) Find four consecutive terms in an A.P. whose sum is 36 and the product of the 2nd
and the 4th is 105. The terms are in ascending order.
Answer should be 3,7,11,15
plz it's urgent​

Answer 1

Answer:

numbers are 3, 7, 11, 15

Step-by-step explanation:

let the 4 terms be a - 3d, a-d, a+d and a+3d

Sum = 4a = 36, so a = 9

Product of 2nd and 4th = (9-d)(9+3d)=105

81+27d-9d-3d2 = 105

3d2-18d+24=0

d2-6d+8=0

(d-2)(d-4)=0

so d=2, or 4

numbers are

3, 7, 11, 15

or

-3, 5, 13, 21

Answer 2

Answer:

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

Step-by-step explanation:

S 4 =36

⇒4/2 [2a+(4−1)d]=36

⇒2a+3d=18⟶(1)

A/Q, a2 x a3 = a1 x a4 + 8

⇒(a+d)(a+2d)=a(a+3d)+8

⇒d^2=4

⇒d=±2

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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