four numbers are in AP whose sum is 40 and the ratio of the product of first and fourth term II II and III term is 2 is to 3 find the number
Answers
Step-by-step explanation:
let us consider the terms of AP
are (a - 3d),(a - d) ,(a + d), (a + 3d)
there sum is 40
(a - 3d )+ (a - d)+ (a + d) + (a + 3d)
= 40
a - 3d + a - d + a + d + a + 3d = 40
→ 4a = 40
→ a = 40 /4
→ a = 10
the ratio of the product of first and fourth term is 2 : 3
→ (a - 3d) / ( a + 3d) = 2 /3
→ (a - 3d) = ( 2/3 ) (a + 3d)
→ (a - 3d) = (2a + 6d) / 3
put the value of a
→ 3( 10 - 3d) = ( 2(10) + 6d)
→ 30 - 9d = 20 + 6d
→ -9d - 6d = 20 - 30
→ -15d = -10
→ d = 2 /3
therefor the terms are
( a - 3d) = ( 10 - 3 (2/3))
= ( 10 - 2)
= 8
( a - d) = ( 10 - (2/3))
= ( 30 - 2) / 3
= 28 / 3
( a + d) = ( 10 + (2/3))
= ( (30 + 2) / 3)
= 32 / 3
(a + 3d ) = ( 10 + 3( 2/ 3))
= ( 10 + 2)
= 12
The numbers are 8 , (28 / 3 ) ,
(32 /3) and 12
» Four numbers are in AP whose sum is 40.
• Let four numbers be (a - 3d), (a - d), (a + d), (a + 3d)
A.T.Q.
=> a - 3d + a - d + a + d + a + 3d = 40
=> 4a = 40
=> a = 40/4
=> a = 10 _______ (eq 1)
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» The ratio of the product of first and fourth term is 2:3.
Cross-multiply them..
=> 3(a - 3d) = 2(a + 3d)
=> 3a - 9d = 2a + 6d
=> 3(10) - 9d = 2(10) + 6d [From (eq 1)]
=> 30 - 9d = 20 + 6d
=> 30 - 20 = 6d + 9d
=> 10 = 15d
=> d = 10/15
=> d = 2/3 ________ (eq 2)
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» We have to find the numbers. (Means AP.)
• a - 3d = 10 - 3(2/3)
=> 10 - 2 = 8
• a - d = 10 - 2/3
=> (30 - 2)/3 = 28/3
• a + d = 10 + 2/3
=> (30 + 2)/3 = 32/3
• a + 3d = 10 + 3(2/3)
=> 10 + 2 = 12
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__________ [ANSWER]
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✡ From above calculations we get numbers 8, 28/3, 32/3 and 12
Add all the numbers.
=> 8 + 28/3 + 32/3 + 12
=> (24 + 28 + 32 + 28)3
=> 120/3
=> 40
_______ [HENCE VERIFIED]
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