The sum of 3 term of an ap is 21 and the product of the first and third term exceed the second term by 6 , find 3 terms
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Answered by
304
Let the three terms of AP are a,a+d,a+2d.
Now,
As per the question ,
a+a+d+a+2d = 3a+3d = 21
i.e. a+d = 7 or, d= 7-a.
Now,
a(a+2d)-(a+d) = 6
a{a+2(7-a)}-7 = 6
a{a+14-2a}=13
a{14-a} =13
14a-a^2 =13
a^2 -14a +13 = 0
By applying quadratic formula, you will get
a= 13, 1.
d will be 7-a i.e. -6 ,6.
Now, there will be two AP
13,7, 1 and 1,7,13.
Now,
As per the question ,
a+a+d+a+2d = 3a+3d = 21
i.e. a+d = 7 or, d= 7-a.
Now,
a(a+2d)-(a+d) = 6
a{a+2(7-a)}-7 = 6
a{a+14-2a}=13
a{14-a} =13
14a-a^2 =13
a^2 -14a +13 = 0
By applying quadratic formula, you will get
a= 13, 1.
d will be 7-a i.e. -6 ,6.
Now, there will be two AP
13,7, 1 and 1,7,13.
Answered by
172
hey mate here is your answer
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