问题1.在GP中找到三个数字，它们的总和为65，乘积为3375。
解决方案：
Let the three numbers in G.P as a/r, a, ar.
Given that product of three numbers is 3375.
Therefore,
a/r * a * ar = 3375
a3 = 3375
a3 = (15)3
Also, a/r + a + ar = 65
a(1/r + 1 + r) = 65
Substituting a = 15
15 (1/r + 1 + r) = 65
1/r + 1 + r = 65/15
1/r + 1 + r = 13/3
3r2 – 10r + 3 = 0
3r2 – 9r – r + 3 = 0
3r(r-3) – (r-3) = 0
(r-3)(3r-1) = 0
r = 3,1/3.
On taking r = 3, we get
a/r = 15 / 3 = 5 , a = 15 , ar = 15 * 3 = 45
On taking r = 1/3, we get
a/r = 15 * 3 = 45, a = 15, ar = 15 * 1/3 = 5
Therefore, the three terms are 5, 15, 45 or 45, 15, 5.
问题2。在GP中找出三个数字，它们的总和为38，其乘积为1728。
解决方案：
Let the three numbers in G.P as a/r, a, ar.
Given that product of three numbers is 1728.
Therefore,
a/r * a * ar = 1728
a3 = 1728
a3 = (12)3
Also, a/r + a + ar = 38
a (1 /r + 1 + r) = 38
Substituting a = 12
12 (1/r + 1 + r) = 38
1/r + 1 + r = 38/12
1/r + 1 + r = 19/6
6r2 – 13r + 6 = 0
6r2 – 9r – 4r + 6 = 0
3r(2r-3) – 2(2r-3) = 0
(2r-3)(3r-2) = 0
r = 3/2, 2/3.
On taking r = 3, we get
a/r = 12 * 2/3 = 8, a = 12, ar = 12 * 3/2 = 18
On taking r = 1/3, we get
a/r = 12 * 3/2 = 18 , a = 12, ar = 15 * 2/3 = 8
Therefore, the three terms are 8, 12, 18 or 18, 12, 8.
问题3. GP的前三个项的总和为13/12，其乘积为-1。找到GP
解决方案：
Let the three numbers in G.P as a/r, a, ar.
Given that product of three numbers is -1.
Therefore, a/r * a * ar = -1
a3 = -1
a3 = (-1)3
a = -1
Also, a/r + a + ar = 13/12
a (1 /r + 1 + r) = 13/12
Substituting a = -1
-1 (1/r + 1 + r) = 13/12
1/r + 1 + r = -13/12
12r2 + 25r + 12 = 0
12r2 + 16r + 9r + 12 = 0
4r(3r+4) + 3(3r+4) = 0
(4r+3)(3r+4) = 0
r = -3/4, -4/3.
On taking r = -3/4, we get
a/r = -1 * -4/3 = 4/3, a = -1, ar = -1 * -3/4 = 3/4
On taking r = -4/3, we get
a/r = -1 * -3/4 = 3/4, a = -1, ar = -1 * -4/3 = 4/3
Therefore, the three terms are 4/3, -1, 3/4 or 3/4, -1, 4/3.
问题4. GP中三个数字的乘积为125，而它们的成对乘积之和为175/2。找到他们。
解决方案：
Let the three numbers in G.P as a/r, a, ar.
Given that product of three numbers is 125.
Therefore, a/r * a * ar = 125.
a3 = 125
a3 = (5)3
a = 5
Also,
a/r * a + a/r * ar + a*ar = 175/2
a2 (1 /r + 1 + r) = 175/2
Substituting a = 5
25 (1/r + 1 + r) = 175/2
1/r + 1 + r = 7/2
2r2 -5r + 2 = 0
2r2 -4r – r + 2= 0
2r(r – 2) – (r – 2) = 0
(2r-1)(r-2) = 0
r = 2, 1/2.
On taking r = 2, we get
a/r = 5/2, a = 2, ar = 5 * 2 = 10
On taking r = 1/2, we get
a/r = 5 * 2 = 10, a = 5, ar = 5 * 1/2 = 5/2
Therefore, the three terms are 5/2, 5, 10 or 10, 5, 5/2.
问题5. GP的前三个项的总和为39/10，其乘积为1。求出公共比率和项。
解决方案：
Assume the three numbers in G.P as a/r, a, ar.
Given that product of three numbers is 1.
Therefore, a/r * a * ar = 1
a3 = 1
a3 = (1)3
Also, a/r + a + ar = 39/10
a (1 /r + 1 + r) = 39/10
Substituting, a = 1,
1 (1/r + 1 + r) = 39/10
1/r + 1 + r = 39/10
10r2 – 29r + 10 = 0
10r2 – 25r – 4r + 10 = 0
5r(2r-5) – 2(2r-5) = 0
(5r-2)(2r-5) = 0
r = 2/5, 5/2.
On taking r = 2/5, we get
a/r = 1 * 5/2 = 5/2, a = 1, ar = 1 * 2/5 = 2/5
On taking r = 5/2, we get
a/r = -1 * 2/5 = 2/5, a = 1, ar = 1 * 5/2 = 5/2
Therefore, the three terms are 5/2, 1, 2/5 or 2/5, 1, 5/2.
问题6. GP中三个数字的总和为14。如果前两个项分别增加1，而第三个项减少1，则结果数在AP中查找。
解决方案：
Let the three numbers in G.P as a/r, a, ar.
First two terms are increased by 1 and third term is decreased by 1 , then it becomes A.P
a/r + 1, a + 1, ar – 1 is an A.P
Therefore,
ar – 1 – a – 1 = a – 1 – a/r – 1
ar – a – 2 = a – a/r
ar + a/r = 2a + 2 —(1)
Since, a/r + a + ar = 14
Replacing a/r + ar = 14 – a in equation (1)
14 – a = 2a + 2
12 = 3a
a = 4
Substituting a in equation (1)
r + 1/r = 5/2
2r2 – 5r + 2 = 0
2r2 – 4r -r + 2 = 0
2r(r-2) -(r-2) = 0
(r-2)(2r-1) = 0
r = 2, 1/2
On taking r = 2,
a/r = 4/2 = 2, a = 4, ar = 4*2 = 8
On taking r = 1/2
a/r = 4*2 = 8, a = 4 , ar = 4/2 = 2
Hence, the numbers are 2, 4, 8 or 8, 4, 2.
问题7. GP中三个数字的乘积为216。如果将2、8、6加到它们上，则结果在AP中。查找数字。
解决方案：
Let the three numbers be a/r, a, ar
Given that product of three number is 216.
Therefore,
a/r * a * ar = 216
a3 = (6)3
a = 6
Adding 2 on a/r, 8 on a and 6 on ar, it becomes an A.P
Therefore,
ar + 6 – a – 8 = a + 8 – a/r – 2,
Substituting a = 6
6r2 – 20r + 6 = 0
6r2 – 18r -2r + 6 = 0
6r(r-3) -2(r-3) = 0
r = 3, 1/3
On taking r = 3
a/r = 6/3 = 2, a = 6, ar = 6*3 = 18
On taking r = 1/3
a/r = 6*3 = 18, a = 6, ar = 6* 1/3 = 2
Hence, the numbers are 2, 6, 18 or 18, 6, 2.
问题8.在GP中找到三个数字，它们的乘积为729，而其乘积对的总和为819。
解决方案：
Let the three number be a/r, a, ar.
Given that product of three number is 729.
Therefore,
a/r * a * ar = 729
a3 = (9)3
a = 9
Also, sum of their products in pairs is 819.
a/r * a + a/r * ar + a * ar = 819
(a)2 * (1/r + 1 + r) = 819
Substituting a = 9
81 * (1/r + 1 + r) = 819
(1/r + 1 + r) = 91/9
9r2 – 82r + 9 = 0
9r2 – 81r – r + 9 = 0
9r(r-9) – (r-9) = 0
(r-9) (9r-1) = 0
r = 9, 1/9
On taking r = 9
a/r = 9/9 = 1, a = 9, ar = 9 * 9 = 81
On taking r = 1/9
a/r = 9 * 9 = 81, a = 9, ar = 9 * 1/9 = 1
Hence, the numbers are 1, 9, 81 or 81, 9, 1.
问题9. GP中三个数字的总和为21，其平方和为189。找到数字。
解决方案：
Let the three numbers be a/r, a, ar
Given that a/r + a + ar = 21 —(1) and,
(a/r)2 + (a)2 + (ar)2 = 189 —(2)
We know that,
(A + B + C)2 = A2 + B2 +C2 + 2(AB + BC + CA)
Replacing A = a/r, B = a, C = ar,
21*21 = 189 + 2(a/r * a + a* ar + a*a)
126 = a2 [ 1/r + r + 1 ] —-(3)
From (1) we have 21 = a [ 1/r + r + 1] —(4)
Dividing (3) by (4), we get
a = 6
Substituting a in equation (4)
1/r + r + 1 = 7/2
2r2 – 5r + 2 = 0
2r2 – 4r – r + 2 = 0
2r(r-2) – (r-2) = 0
r = 2, 1/2
On taking r = 2
a/r = 3, a = 6, ar = 12
On taking r = 1/2
a/r = 12, a = 6, ar = 3
Hence, the numbers are 3, 6, 12 or 12, 6, 3