Question

The sum of 3 term of an Ap is 21 and product of first and 3 third term excees the second term twice to find the 3 numbers ​

Answer 1

Let the first three term of AP are a-d ,a , a+d

Given sum of first three term is 33

therefore,a+a-d +a+d =33

3a=33

a=11

Product of first and third exceeds 29 by second term:-

= (a-d)(a+d) =a+29( put the value of a)

=11²-d²=40

=81=d²

= d=±9

Series are 20,11,2………, Or. 2,11,20….,,

Answer 2

Step-by-step explanation:

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