# In an AP the 3rd term is equal to 5 times the 6th term . The 2nd term is lesser by 3 than twice the 4th term . Find the terms of AP

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by Saadhana

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by Saadhana

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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 . . .

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 . . .