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Question
Divide 32 into four parts which are the four terms of an AP such that the product of the first and fourth terms is to product of the second and the third terms as 7:15.
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Solution
Let the four parts in AP be (a-3d) ,(a-d) ,(a+d) and (a+3d) Then.
(a - 3d) + (a - d) + (a + d) + (a + 3d) = 32
⇒ 4a = 32
⇒ a= 8 ................(1)
Also,
(a-3d) (a+3d) : (a-d) (a+d) = 7:15
`⇒((8-3"d") (8+3"d") )/((8-"d")(8+"d") ) = 7/15` [From (1)]
⇒`(64-9"d"^2)/(64-"d"^2) = 7/15`
⇒`15 (64-9"d"^2) = 7(64 - "d"^2)`
⇒`960+135"d"^2 = 448 - 7"d"^2`
⇒`135 "d"^2 - 7"d"^2 = 960-448`
⇒`128"d"^2 = 512`
⇒`"d"^2=4`
⇒`"d" = +-2`
When a = 8 and d = 2,
a - 3d =8-3×2=8-6=2
a - d =8-2=6
a+d = 8+2 =10
a+3d =8+3×2=8+6=14
When a = 8 and d = -2 ,
a-3d = 8 - 3× (-2) = 8 +6=14
a-d = 8 - (- 2) = 8 + 2 = 10
a+d = 8 - 2 = 6
a+3d = 8 + 3 ×(- 2) = 8 - 6 = 2
Hence, the four parts are 2, 6, 10 and 14.
