Question

an ap is formed by the multiple of 4 between 50 and 350 find the sum of the next eight terms of AP​

Answer 1

Step-by-step explanation:

Let the 4 consecutive numbers in AP be

(a−3d),(a−d),(a+d),(a+3d)

According to the question,

a−3d+a−d+a+d+a+3d=32

4a=32

a=8

Now,

(a−d)(a+d)

(a−3d)(a+3d)

=

15

7

15(a

2

−9d

2

)=7(a

2

−d

2

)

15a

2

−135d

2

=7a

2

−7d

2

8a

2

=128d

2

Putting the value of a, we get,

d

2

=4

d=±2

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