Question

four terms in ap have sum 28 product of the extreme terms and that of the middle term are in the ratio 5:6 find the largest term​

Answer 1

answer : largest term = 10

explanation : let (a - 3d), (a - d) , (a + d), (a + 3d) four terms in AP, where a and d are two constant terms.

a/c to question,

sum of four terms = 28

⇒ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 28

⇒ 4a = 28

⇒ a = 7 .......(1)

again,

{(a - 3d)(a + 3d)}/{(a - d)(a + d)} = 5/6

⇒(a² - 9d²)/(a² - d²) = 5/6

⇒6(a² - 9d²) = 5(a² - d²)

⇒ 6a² - 54d² = 5a² - 5d²

⇒6a² - 5a² = 54d² - 5d²

⇒a² = 49d²

⇒a = ±7d

from equation (1), 7 = ±7d ⇒d = ±1

[you can take either 1 or -1 as common difference , you will get the same result ]

all four terms are ;

(a - 3d) = 7 - 3(1) = 4

(a - d) = 7 - 1 = 6

(a + d) = 7 + 1 = 8

(a + 3d) = 7 + 3(1) = 10

so, largest term = 10

Answer 2

Answer:

Largest Term = 10

Step-by-step explanation:

Let say Four terms are

a-3d , a - d , a+d , a + 3d

a -3d is first term

2d is common difference

Sum = a - 3d + a - d + a + d + a + 3d = 28

=> 4a = 28

=> a = 7

product of the extreme terms and that of the middle terms are in the ratio 5:6

=> (a - 3d)(a + 3d) / (a - d)(a + d)  = 5/6

=> (a² - 9d²)*6 = 5(a² - d²)

=> 6a² - 54d² = 5a² - 5d²

=> a² = 49d²

=> a = ± 7d

a = 7

=> d = ± 1

Terms are

4 , 6 , 8 , 10

or

10 , 8 , 6 , 4

so Largest Term = 10

Verification : 4 + 6 + 8 + 10 = 28

4 * 10 /(6 * 8)  = 10/(6 * 2)  = 5/6