Asked by jude
The sum of the three numbers in Arithmetic Progression is 33. If the numbers are increased by 2, 1, and 6 respectively the new numbers will be in Geometric progression. Find these numbers.
Answers
Answered by
Reiny
let the 3 numbers in AP be
a-d, a, and a+d
a-d + a + a-d = 33
3a = 33
a = 11
after the changes, the new numbers are:
a-d + 2, a+1, and a+d+6
or
13-d , 12, and 17+d
which are now is GP, so
12/(13-d) = (17+d)/12
144 = 221 - 4d - d^2
d^2 + 4d - 77 = 0
(d + 11)(d-7) = 0
d = -11 or d = 7
case 1: a = 11 , d = -11
the numbers are
22, 11, and 0
case 2: a = 11, d = 7
the numbers are
4, 11, 18
checking the last one:
4, 11, and 18 are in AP
if we do the changes, the new numbers are:
6, 12, and 24
is 12/6 = 24/12 ? ----> YES
check the first one in the same way
a-d, a, and a+d
a-d + a + a-d = 33
3a = 33
a = 11
after the changes, the new numbers are:
a-d + 2, a+1, and a+d+6
or
13-d , 12, and 17+d
which are now is GP, so
12/(13-d) = (17+d)/12
144 = 221 - 4d - d^2
d^2 + 4d - 77 = 0
(d + 11)(d-7) = 0
d = -11 or d = 7
case 1: a = 11 , d = -11
the numbers are
22, 11, and 0
case 2: a = 11, d = 7
the numbers are
4, 11, 18
checking the last one:
4, 11, and 18 are in AP
if we do the changes, the new numbers are:
6, 12, and 24
is 12/6 = 24/12 ? ----> YES
check the first one in the same way
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