Question

solve (a+6d)-(a-3d)=15 so that after solving this comes 3d=15​

Answer 1

Answer:

Let the 4 consecutive numbers in AP be

(a-3d), (a-d), (a+d) , (a+3d)(a−3d),(a−d),(a+d),(a+3d)

According to the question,

a-3d+a-d+a+d+a+3d=32a−3d+a−d+a+d+a+3d=32

4a=324a=32

a=8a=8

Now,

\dfrac{(a-3d)(a+3d)}{(a-d)(a+d)} = \dfrac{7}{15}

(a−d)(a+d)

(a−3d)(a+3d)

=

15

7

15(a^{2}-9d^{2})=7(a^{2}-d^{2})15(a

2

−9d

2

)=7(a

2

−d

2

)

15a^{2}-135d^{2}=7a^{2}-7d^{2}15a

2

−135d

2

=7a

2

−7d

2

8a^{2}=128d^{2}8a

2

=128d

2

Putting the value of a, we get,

d^{2}=4d

2

=4

d=\pm2d=±2

So, the four consecutive numbers are 2,6,10,142,6,10,14 or 14,10,6,214,10,6,2.

Answer 2

Step-by-step explanation:

(a+6d)-(a-3d) = 15

any how it can' t come 3d=15

if the que would be like this

(a+6d)-(a+3d)=15

then,

a+6d-a-3d=15

a-a+6d-3d=15

0+3d=15

3d=15

then the solution of the que can be that