Math, asked by niteshshaw723, 6 months ago

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Answered by nabilahabubaker
1

In case of 4 terms, we usually take 2d as the common difference and the average of the two middle terms should be a.

Therefore, Let a-3d, a-d, a+d and a+3d be the 4 terms.

Now, By Question, we have

(a-3d) + (a-d) + (a+d) + (a+3d) = 32

=> 4a = 32

=> a = 8

Again, By Question, we have

(a-3d)*(a+3d) : (a-d)*(a+d) = 7 : 15

=> a^2 - 9d^2 : a^2 - d^2. = 7 : 15

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

=>15a^2 - 135d^2. = 7a^2 - 7d^2

=>15a^2 - 7a^2. = 135d^2 - 7d^2

=>8a^2. = 128d^2

=>8(8)^2. = 128d^2 (Substituting the value of a)

=>512. = 128d^2

=>4. = d^2

=>d =(+)(-)2

Therefore Common Difference = 2d = (+)(-)4

Therefore, The required AP is -

a-3d = 8 - 3(2) = 2. Or. a-3d = 8-3(-2) = 14

a-d. = 8 - 2. = 6. Or. a-d. = 8-(-2) = 10

a+d. = 8+2. = 10. Or. a+d. = 8+(-2) = 6

a+3d = 8+3(2) = 14. Or. a+3d = 8+3(-2) = 2

Answer:

Step-by-step explanation:thats  the answer your welcome

Answered by karushingneyngp
2

Answer:

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