Question

Pls friends answer the above question ​

Answer 1

Answer:

2,6,10,14 (or) 14,10,6,2

Step-by-step explanation:

Let the four consecutive numbers be (a - 3d), (a - d), (a + d), (a + 3d).

(i) Sum of four is 32:

⇒ a - 3d + a - d + a + d + a + 3d = 32

⇒ 4a = 32

⇒ a = 8.

(ii) Ratio of product of 1st and last term to extremes is 7:15

⇒ (a - 3d)(a + 3d)/(a - d)(a + d) = 7/15

⇒ 15(a - 3d)(a + 3d) = 7(a - d)(a + d)

⇒ 15a² - 135d² = 7a² - 7d²

⇒ 8a² = 128d²

⇒ 8(8)² = 128d²

⇒ 512 = 128d²

⇒ d² = 4

⇒ d = ±2

When a = 8, d = 2:

a - 3d = 2

a - d = 6

a + d = 10

a + 3d = 14

When a = 8, d = -2:

a - 3d = 14

a - d = 10

a + d = 6

a + 3d = 2

Therefore, the numbers are 2,6,10,14 (or) 14,10,6,2

Hope it helps!

Answer 2

HERE is your answer

Let the four consecutive terms of an A.P be a-3d, a-d, a+d and a+3d.

where = >>

extreme terms--> a-3d and a+3d.

mean terms--> a-d and a+d.

Note a trick: When we need to assume more than 2 consecutive terms then we take the consecutive terms in A.P in such a way that on sum the all (d)s eliminate.

eg:Three consecutive terms=

a ,a-d and a+d.

Now,

sum of four consecutive terms = 32

=>a-3d+a-d+ a+d+ a+3d = 32

=>4a = 32

=>a =8.

Now, according to the question-->

(a-3d)(a+3d)/(a-d)(a+d) = 7/15

=>a^2 - 9d^2/a^2 - d^2 = 7/15

{using identities(a-b)(a+b)= a^2 - b^2)

=>64- 9d^2/64 - d^2 = 7/15

now, by cross multiplication---->

=>15(64-9d^2) = 7(64- d^2)

=> 960 - 135 d^2 = 448 - 7 d^2

=>960 - 448 = -7d^2 + 135d^2

=>512 = 128d^2

=>d^2 = 4

=>d = plus minus 2

So,

when d = -2 and a = 8

the numbers are 14,12,10 and 8.

when d = 2 and a = 8

the numbers are 8,10,12 and 14.

thanks