Question

DIVIDE 72 IN FOUR SUCH PARTS THAT THEY ARE IN AP AND THE RATIO IN THE PRODUCT OF FIRST AND THIRD PARTS TO THEY PRODUCT OF THE SECOND AND FOURTH PARTS IS 5RATIO 8

Answer 1

12, 16 , 20 and 24 or, 180, 72, -36 , -144 are parts of 72 in AP.

Divide 72 in four such parts that they are in AP and the ratio of the product of first and third parts to the product of second and fourth parts is 5 : 8.

Let a - 3d , a - d , a + d, a + 3d are four parts of 72 in AP.

⇒(a - 3d) + (a - d) + (a + d) + (a - 3d) = 72

⇒4a = 72

⇒a = 18

so,

• 1st term = 18 - 3d
• 2nd term = 18 - d
• 3rd term = 18 + d
• 4th term = 18 + 3d

a/c to question,

(1st × 3rd)/(2nd × 4th) = 5/8

⇒{(18 - 3d)(18 + d)}/{(18 - d)(18 + 3d)} = 5/8

⇒8(18 - 3d)(18 + d) = 5(18 - d)(18 + 3d)

⇒8[18(18 + d) - 3d(18 + d)] = 5[18(18 + 3d) - d(18 + 3d)]

⇒8[324 + 18d - 54d - 3d²] = 5[324 + 54d - 18d - 3d²]

⇒3 × 324 - 8 × 36d - 5 × 36d - 8 × 3d² + 5 × 3d² = 0

⇒972 - 13 × 36d - 9d² = 0

⇒d² + 52d - 108 = 0

⇒d² + 54d - 2d - 108 = 0

⇒d(d + 54) - 2(d + 54) = 0

⇒(d + 54)(d - 2) = 0

⇒d = -54 , 2

now numbers are if d = 2

18 - 3d = 18 - 6 = 12

18 - d = 18 - 2 = 16

18 + d = 18 + 2 = 20

18 + 3d = 18 + 6 = 24

hence numbers are 12, 16, 20, 24.

numbers if d = -54

18 - 3d = 18 - 3(-54) = 18 + 162 = 180

18 - d = 18 - (-54) = 18 + 54 = 72

18 + d = 18 - 54 = -36

18 + 3d = 18 + 3d = 18 - 162 = -144

hence numbers are 180, 72 , -36 , -144.

both groups of numbers are correct.

Therefore 12, 16 , 20 and 24 or, 180, 72 , -36 and -144 are parts of 72.

Answer 2

Divide the number $72$ in the AP of four parts.

Explanation:

• Let the four terms in the A.P be $a_1,\ a_2,\ a_3,\ and\ a_4$.
• Now we have, $a_1+a_2+a_3+a_4=72$  
• From the formula of the sum an A.P we get, $\frac{4}{2} (2a_1+3d)=72\\->2a_1+3d=36\ \ \ \ \ \ \ \ \ \ \ ----(a)$  
• Now it is required that the ratio of the products of first-third terms and second-fourth terms is $5:8$.
• Hence we get,          
• $\frac{a_1a_3}{a_2a_4}=\frac{5}{8}\\ ->8a_1(a_1+2d)=5(a_1+d)(a_1+3d)\\ ->8a_1^2+16a_1d=5a_1^2+20a_1d+15d^2\\->3a_1^2-4a_1d-15d^2=0\\->(3a_1+5d)(3d-a_1)=0 \\->a_1=\frac{-5d}{3} ,\ \ a_1=3d$      ----(b)  
• From (a) and (b) we get two possible AP,
• $a_1=3d,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_1=\frac{-5d}{3} \\-> 9d=36\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ->-10d+6d=72\\->d=4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ->d=18\\AP_1-> 12, 16, 20, 24\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ AP_2-> -30, -12, 6, 24$    
• Assuming the required parts are to be of positive numbers the four parts are $12,16,20,24$.