The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:15. Find the numbers
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Narendra931:
because if you assume these than you can solv easely
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Let the no/s be of the form (a-3d) (a-d) (a+d) (a+2d)
Then according to question,
(a-3d)+(a-d)+(a+d)+(a+3d) = 32
4a = 32
a = 8
Also,
=
15(a²-9d²) = 7(a²-d²)
15a² - 135d² = 7a² - 7d²
8a² - 128d² = 0
⇒Divide by 8 on whole
⇒ a² - 16d²
⇒Substituting a =8, Then
⇒ 8² - 16d²
⇒64 - 16d²
⇒(8+4d)(8-4d)
Then d = +2, -2
______________________________________________________
If d = +2 , Then
No/s - (2, 6, 10, 14)
______________________________________________________
If d = -2 , Then
No/s - (14,10, 6, 2)
______________________________________________________
Hence the no/s would be (2, 6, 10, 14) or (14, 10, 6, 2)
☺☺☺ Hope this Helps ☺☺☺
Then according to question,
(a-3d)+(a-d)+(a+d)+(a+3d) = 32
4a = 32
a = 8
Also,
=
15(a²-9d²) = 7(a²-d²)
15a² - 135d² = 7a² - 7d²
8a² - 128d² = 0
⇒Divide by 8 on whole
⇒ a² - 16d²
⇒Substituting a =8, Then
⇒ 8² - 16d²
⇒64 - 16d²
⇒(8+4d)(8-4d)
Then d = +2, -2
______________________________________________________
If d = +2 , Then
No/s - (2, 6, 10, 14)
______________________________________________________
If d = -2 , Then
No/s - (14,10, 6, 2)
______________________________________________________
Hence the no/s would be (2, 6, 10, 14) or (14, 10, 6, 2)
☺☺☺ Hope this Helps ☺☺☺
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