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2015-10-13T22:18:05+05:30

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Let a be the 1st term and d be the common difference

using nth term = a + (n - 1)d                   {where n is the number of terms

according to the question 

5[a + (6-1)d] = a + (3-1)d

⇒ 5a + 25d = a + 2d

⇒ 4a + 23d = 0          ...................... ( i )

Again 

a + (2 - 1)d + 3 = 2[a + (4-1)d]

⇒ a + d + 3= 2a + 6d 

⇒ a + 5d - 3 = 0         ......................... ( ii )


∴ 4a + 23d = 0       ...................... ( i )

and 

∴ a + 5d - 3 = 0      ........................( ii )

now  ( i ) - 4 x ( ii ) 

∴  4a + 23d - 4a - 20d + 12 = 0

⇒ d = - 4 

putting d in ( i) we get 

 ∴ 4a + 23 x -4 = 0 

⇒  a = 23 

so the terms of AP 

23 , 19 , 15 , 11 . . . 


3 5 3
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