Three numbers form an arithmetic sequence having a common difference of 4. If the first number is increased by 2, the second number by 3, and the 3rd number by 5, the resulting numbers form a geometric sequence. How do you find the original numbers?

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Answer 1 · 2016-05-01T19:18:13

#23, 27, 31#

The original sequence is:

#a_1 = a#

#a_2 = a+4#

#a_3 = a+8#

The modified sequence is:

#b_1 = a+2#

#b_2 = a+7#

#b_3 = a+13#

Since #b_1, b_2, b_3# is a geometric sequence, the middle term #b_2# must be a geometric mean of #b_1# and #b_3#. Hence:

#b_2^2 = b_1 b_3#

#a^2+14a+49 = (a+7)^2 = (a+2)(a+13) = a^2+15a+26#

Subtracting #a^2+14a+26# from both ends, we find:

#a=23#

So #a_1, a_2, a_3 = 23, 27, 31#