Question

Q15. The first of four given numbers are in G.P , and their last three are in A.P . with common difference 6. If first and fourth numbers are equal, then the first number is:

1 point

2

4

6

8

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Answer 1

Answer:

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 a² −12a+36=a² +6a

or 18a=36

∴a=2.

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

Step-by-step explanation:

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