Question

What is the common difference of four terms in an A.P. such that the ratio of the product of the first and fourth terms to that of the second
and third is 2 : 3 and the sum of all four terms is 20

a. 3
b. 1
c. 4
d. 2

plz....help out..​

Answer 1

Answer:

2

Step-by-step explanation:

Sure,

Let the terms in A.P be

(a-3d), (a-d), (a+d), (a+3d)

according to the condition,

there sum is 20

therefore,

(a-3d)+(a-d)+(a+d)+(a+3d) = 20

4a = 20

a = 20/4 = 5

a = 5 ...(1)

From next condition

$\frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{2}{3}$

$\frac{ {a}^{2} - 9 {d}^{2} }{ {a}^{2} - {d}^{2} } = \frac{2}{3}$

since, (a+b)(a-b) = (a²-b²)

inserting the value of a from equation (1)

We get ,

3(25-9d²) = 2(25-d²)

75 - 27d² = 50 - 2d²

25 = 25d²

d² = 1

d = 1 .....(2)

now,

(a-3d) = 5-3 = 2

(a-d) = 5-1 = 4

(a+d) 5+1 = 6

(a+3d) = 5+3 = 8

therefore the terms in the arithmetic progression are,,

2 , 4, 6 , 8

and actual difference between the terms is 2

if you think I was right surely mark it as BRAINLIEST ANSWER