Question

Find out four numbers such that, first three numbers are in G.P., last three numbers are in A.P. having common difference 6, first and last numbers are same.
A)8, -4, -2, -8
B)8, -4, 2, 8
C)8, 4, 2, 8
D)8, 4, -2, 8​

Answer 1

Answer:

B) 8, -4, 2, 8

Step-by-step explanation:

Last three of the four numbers are in A.P. and hence they may be chosen as a−d,a,a+d

Also the first number is same as the last one i.e. a+d.

Therefore the four numbers are a+d,a−d,a,a+d. The first three of the above four are in G.P.

∴(a−d) 2 =a(a+d). But d=6 given

∴(a−6) 2 =a(a+6)

or a 2 −12a+36=a 2 +6a

or 18a=36 ∴a=2.

Putting for a and d the four numbers are 8,−4,2,8, which satisfy the given conditions.

Answer 2

Answer:

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