Question

What is the common difference of four terms in A.P.
such that the ratio of the product of the first fourth
term to that of the second and third term is 2:3 and
the sum of all four terms is 20?
(a) 3
(b) 1
(c) 4
(d) 2

$please \: answer \: with \: explanation$
​

Answer 1

Answer:

2

Step-by-step explanation:

Let four consecutive terms ; (a - 3d), (a - d) , (a + d), (a + 3d) are in AP.

a/c to question,

sum of all four terms is 20

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

or, 4a = 20

or, a = 5......(1)

again a/c to question,

the ratio of the product of first and fourth terms to that of second term and third is 2:3.

so, (a - 3d)(a + 3d)/(a - d)(a + d) = 2/3

or, (a² - 9d²)/(a² - d²) = 2/3

or, 3a² - 27d² = 2a² - 2d²

or, a² = 25d²

from equation (1),

(5)² = 25d²

or, d² = 1 => d = ±1

here common difference = (a - d) - (a - 3d) = (a + d) - (a - d) = (a + 3d) - (a + d) = 2d

so, common difference = 2(±1) = ±2

Answer 2

Answer:

options no (d) ..... 2

Step-by-step explanation:

Let four consecutive term; (a - 3d),( a - d)(a + d),(a + 3d)

A.T.Q

sum of all four terms is 20

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

4a = 20

a = 5...........(1)

Again A.T.Q

The ratio of the product of first and fourth to that of second term and third is 2:3

so,

(a - 3d)(a + 3d)/( a - d)(a + d) = 2/3

(a² -9d²)/(a² - d²) = 2/3

3a² - 27d² = 2a² - 2d²

a² = 25d²

from equation....(1),

(5)² = 25d²

d² = 1

d = +, - 1

common different = ( a - d) - (a - 3d) = ( a - d) - (a + d) = (a + 3d) - (a + d) = 2d

so, common different = 2(-,+ 2)

+,-,2

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