问题1. AP的前三个项的总和为21,并且第一项和第三项的乘积比第二项多6,找到三个项。
解决方案:
We are given that the sum of first three terms are 21. Let’s suppose these three terms are a-d, a, a+d.
So,
a-d + a + a+d = 21
3a = 21
a= 7
And we are given
(a-d)*(a+d) – a = 6
a2 – d2 -a = 6
Putting value of a in this equation we can get value of d.
(7)2 – (d)2 -7 = 6
49 – d2 – 7 = 6
d2 = 36
d= ±6
For d = 6 three terms of AP are 1, 7, 13.
For d = -6 three terms are 13, 7, 1.
问题2. AP中有三个数字。如果这些数字的总和为27,乘积为648,则找到数字。
解决方案:
We are given that the sum of first three terms are 27. Let’s suppose these three terms are a-d, a, a+d.
So
a-d + a + a+d = 27
3a = 27
a= 9
And we are given
(a-d)*a*(a+d) =648
a3 – ad2 = 648
Putting value of a in this equation we can get value of d.
93– 9d2 =648
729- 9d2 =648
9d2 = 81
d2 = 9
d= ±3
For d = 3 three terms of AP are 6, 9, 12.
For d = -3 three terms are 12, 9, 6.
问题3.在AP中找到四个数字,它们的总和是50,最大的数字是最小的4倍。
解决方案:
We are given that the sum of first four terms are 50. Let’s suppose these four terms are a-3d, a-d, a+d, a+3d.
So
a-3d + a-d + a+d + a+3d = 50
4a = 50
a = 25/2
And we are given greatest number is 4 times the least,
4(a-3d) = a +3d
4a – 12d = a + 3d
3a = 15d
a = 5d
Putting value of a, we get
d = a/5 = 5/2
So the four terms of AP are
a – 3d = 25/2- 3*5/2 = 5
a – d = 25/2- 5/2 = 10
a + d = 25/2 + 5/2 = 15
a + 3d = 25/2 + 3*5/2 = 20
问题4. AP中三个数字的总和为12,而多维数据集的总和为288。找到数字。
解决方案:
We are given that the sum of first three terms are 12. Let’s suppose these three terms are a-d, a, a+d.
So
a-d + a + a+d = 12
3a = 12
a= 4
And we are given sum of there cubes is 288.
(a-d)3 + a3 + (a+d)3 = 288
a3-d3-3ad(a-d) +a3+ a3 +d3 + 3ad(a+d) = 288
Putting value of a in this equation,
43-d3-3*4*d(4-d) + 43 + 43 + d3 + 3*4*d(4+d) = 288
64- d3-12d(4-d) + 64+64+ d3+ 12d(4+d) = 288
192+24d2 =288
24d2=96
d2= 4
d= ±2
For d = 2 three terms of AP are 2, 4, 6.
For d = -2 three terms are 6, 4, 2.
问题5.如果AP中三个数字的总和是24,它们的乘积是440。找到数字。
解决方案:
We are given that the sum of first three terms are 24. Let’s suppose these three terms are a-d, a, a+d.
So
a-d + a + a +d = 24
3a = 24
a = 8
And we are given
(a-d)*a*(a+d) =440
a3 – ad2 = 440
Putting value of a in this equation we can get value of d.
83– 8d2 = 440
512- 8d2 = 440
8d2 = 72
d2 = 9
d = ±3
For d = 3 three terms of AP are 5, 8, 11.
For d = -3 three terms are 11, 8, 5.
问题6.四边形的角度在AP中,其共同差为10。找到这些角度。
解决方案:
We are given that the angles of a quadrilateral are in AP and we know that there are four angles in quadrilateral and sum of all angles in 360.
So, let’s suppose these four angles are a-3d, a-d, a+d, a+3d.
We know sum of these angles is 360.
a-3d + a-d + a+d + a+3d = 360
4a = 360
a =90
We are given that common difference is 10.
(a-d) – (a-3d) = 10
2d = 10
d = 5
So the four angles are
a-3d = 90-15= 75
a-d = 90 – 5 = 85
a+d = 90 + 5 = 95
a+3d = 90+15 = 105