Question

If the sum of the first three terms of an AP is twice the sum of the first 6 terms of the same AP, find the fourth term of that AP.

Answer 1

Answer:

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−a2=13

a2−14a+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

Answer 2

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