Question

33. Find 4 consecutive terms in AP whose sum
is 18 and the sum of third and fourth terms
is 19.​

Answer 1

Answer:

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) = 157

15(a 2 −9d 2 )=7(a 2 −d 2 )

15a 2 −135d 2 =7a 2−7d28a

2 =128d 2

Putting the value of a, we get,d 2 =4d=±2

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

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