find 4 numbers in AP such that the sum of 2nd and 3rd terms is 22 & the product of 1st & 4th terms is 85.
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Sum of 2nd and 3rd term :-
a + d + a + 2d = 22
2a + 3d = 22
d = (22 - 2a ) / 3. ...........(1)
the product of 1st and 4th term :
a × ( a + 3d ) = 85
a² + 3ad = 85 .......(2)
From (1) & (2)
a² + 3a × (22 - 2a)/3 = 85
[a² + ( 22a - 2a² ) ] = 85
[ - a² + 22a -85 ] = 0
a² - 22a + 85 = 0
a² - 17a - 5a + 85 = 0
a × ( a - 17 ) - 5 × ( a - 17 ) = 0
( a - 5 ) × ( a - 17 ) = 0
a = 5 , 17
When a = 5
d = (22 - 10) /3
d = 4
A.P is 5 , 9 , 13 , 17
When a = 17
d = ( 22 - 34 ) /3 = -4
A.P is 5 , 1 , -3 , -7
HOPE IT HELPS
Sum of 2nd and 3rd term :-
a + d + a + 2d = 22
2a + 3d = 22
d = (22 - 2a ) / 3. ...........(1)
the product of 1st and 4th term :
a × ( a + 3d ) = 85
a² + 3ad = 85 .......(2)
From (1) & (2)
a² + 3a × (22 - 2a)/3 = 85
[a² + ( 22a - 2a² ) ] = 85
[ - a² + 22a -85 ] = 0
a² - 22a + 85 = 0
a² - 17a - 5a + 85 = 0
a × ( a - 17 ) - 5 × ( a - 17 ) = 0
( a - 5 ) × ( a - 17 ) = 0
a = 5 , 17
When a = 5
d = (22 - 10) /3
d = 4
A.P is 5 , 9 , 13 , 17
When a = 17
d = ( 22 - 34 ) /3 = -4
A.P is 5 , 1 , -3 , -7
HOPE IT HELPS
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