find three consecutive terms of an AP if the sum of the second term and the third term is 130 and the sum of the first and the second term is 90
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Let the three terms of A.P 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 factorising,we get
a=13,1
d=7−a
For a=13,1d=7−13,7−1=−6,6 respectively.
Now, there are two AP's a,a+d,a+2d for two values of a and d
13,7,1 and 1,7,13
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