第 10 类 RD Sharma 解决方案 - 第 9 章算术级数 - 练习 9.5
问题 1:求 x 的值,其中 (8x + 4)、(6x – 2) 和 (2x + 7) 在 AP 中
解决方案:
Since (8x + 4), (6x – 2) and (2x + 7) are in A.P.
Now we know that condition for three number being in A.P.-
⇒2*(Middle Term)=(First Term)+(Last Term)
⇒2(6x-2)=(8x+4)+(2x+7)
⇒12x-4=10x+11
⇒(12x-10x)=11+4
⇒2x=15
⇒x=15/2
Mean value of x=15/2.
问题 2:如果 x + 1、3x 和 4x + 2 在 AP 中,求 x 的值。
解决方案:
Since x + 1, 3x and 4x + 2 are in A.P.
Now we know that condition for three number being in A.P.-
⇒2*(Middle Term)=(First Term)+(Last Term)
⇒2(3x)=(x+1)+(4x+2)
⇒6x=5x+3
⇒x=3
Mean value of x=3.
问题 3:证明 (a – b)²、(a² + b²) 和 (a + b)² 在 AP
解决方案:
We know that condition for three number being in A.P.-
⇒2*(Middle Term)=(First Term)+(Last Term) ———-(1)
First term=(a-b)2
=a2-2ab+b2
Middle term=a2+b2
Last term=(a+b)2
=a2+2ab+b2
Now putting these values in equation(1)-
⇒2(a2+b2)=(a2-2ab+b2)+(a2+2ab+b2)
⇒2(a2+b2)=2a2+2b2
⇒2(a2+b2)=2(a2+b2)
Since L.H.S=R.H.S.
Hence proved.
问题4:AP的三项之和为21,第一项和第三项的乘积超过第二项6,求三项。
解决方案:
Let the first term of the A.P. is=a
and common difference is=d
let three terms are=(a-d),a,(a+d)
Now according to first condition-
⇒(a-d)+a+(a+d)=21
⇒3a=21
⇒a=7
Now according to second condition-
⇒(a-d)(a+d)=a+6
⇒a2-d2=a+6
Putting a=7-
⇒49-d2=7+6
⇒d2=49-13
⇒d2=36
⇒d=6 or d=-6
When a=7 and d=6
First number=a-d=1
Second number=a=7
Third number=a+d=13
When a=7 and d=-6
First number=a-d=13
Second number=a=7
Third number=a+d=1
So required numbers are 1,7 and 13 or 13,7 and 1.
问题 5:AP 中有三个数字 如果这些数字之和是 27 和乘积 648,请找出这些数字。
解决方案:
Let the first term of the A.P. is=a
and common difference is=d
let three terms are=(a-d),a,(a+d)
Now according to first condition-
⇒(a-d)+a+(a+d)=27
⇒3a=27
⇒a=9
Now according to second condition-
⇒(a-d)*a*(a+d)=648
⇒(a2-d2)*a=648
putting a=9-
⇒(81-d2)*9=648
⇒81-d2=648/9
⇒81-d2=72
⇒d2=81-72
⇒d2=9
⇒d=-3 or d=3
When a=9 and d=-3
First term=a-d=12
Second term=a=9
Third term=a+d=6
When a=9 and d=3
First term=a-d=6
Second term=a=9
Third term=a+d=12
means required numbers are 6,9 and 12 or 12,9 and 6.
问题 6:找出 AP 中的四个数,其和为 50,其中最大数是最小数的 4 倍。
解决方案:
Let the first term of the A.P. is=a
and common difference is=d
let three terms are=(a-3d), (a-d), (a+d), (a+3d)
Now according to first condition-
⇒(a-3d)+(a-d)+(a+d)+(a+3d)=50
⇒4a=50
⇒a=50/4
⇒a=25/2
Now according to second condition-
⇒(greatest number)=4*(least number)
⇒a+3d=4(a-3d)
⇒a+3d=4a-12d
⇒3d+12d=4a-a
⇒15d=3a
Dividing by 3-
⇒5d=a
⇒5d=25/2
⇒d=5/2
with the help of a=25/2 and d=5/2-
⇒a-3d=(25/2)-(15/2)=10/2=5
⇒a-d=(25/2)-(5/2)=20/2=10
⇒a+d=(25/2)+(5/2)=30/2=15
⇒a+3d=(25/2)+(15/2)=40/2=20
So required numbers are 5,10,15 and 20.
问题 7:AP 中三个数字的和是 12,它们的立方和是 288。找出这些数字。
解决方案:
Let the first term of the A.P. is=a
and common difference is=d
let three terms are=(a-d),a,(a+d)
Now according to first condition-
⇒(a-d)+a+(a+d)=12
⇒3a=12
⇒a=4
And according to second condition-
⇒(a-d)3+a3+(a+d)3=288
by using (A+B)3=A3+3AB(A+B)+B3
and (A-B)3=A3-3AB(A-B)-B3
⇒{a3-3ad(a-d)-d3}+a3+{a3+3ad(a+d)+d3}=288
⇒3a3-3a2d+3ad2+3a2d+3ad2=288
⇒3a3+6ad2=288
dividing by 3-
⇒a3+2ad2=96
putting a=4-
⇒64+8d2=96
dividing by 8-
⇒8+d2=12
⇒d2=4
⇒d=-2 or d=2
when a=4 and d=-2
a-d=6
a=4
a+d=2
when a=4 and d=2
a-d=2
a=4
a+d=6
so required numbers are 6,4 and 2 or 2,4 and 6.
问题 8:在 AP 中将 56 分为四个部分,使得它们的极值乘积与均值乘积之比为 5 : 6。
解决方案:
Let the first term of the A.P. is=a
and common difference is=d
let required four parts are=(a-3d),(a-d),(a+d),(a+3d)
now since (a-3d),(a-d),(a+d),(a+3d) are parts of 56.
so-
⇒(a-3d)+(a-d)+(a+d)+(a+3d)=56
⇒4a=56
⇒a=14
Now extreme parts are=(a-3d) and (a+3d)
and mean parts are=(a-d) and (a+d)
According to given condition-
⇒{(a-3d)(a+3d)}/{(a-d)(a+d)}=5/6
⇒6{(a-3d)(a+3d)}=5{(a-d)(a+d)}
by using (A-B)(A+B)=A2-B2 –
⇒6{a2-9d2}=5{a2-d2}
putting a=14
⇒6{196-9d2}=5{196-d2}
⇒6*196-54d2=5*196-5d2
⇒6*196-5*196=54d2-5d2
⇒196=49d2
⇒d2=4
⇒d=2 or d=-2
When a=14 and d=2 –
a-3d=8
a-d=12
a+d=16
a+3d=20
When a=14 and d=-2
a-3d=20
a-d=16
a+d=12
a+3d=8
So required parts are 8,12,16 and 20 or 20,16,12 and 8.
问题 9:四边形的角在 AP 中,公差为 10°。找到角度。
解决方案:
Let the angles of quadrilateral are-
(a-3d), (a-d), (a+d), (a+3d)
Now we know that sum of angles in a quadrilateral is=360°
⇒(a-3d)+(a-d)+(a+d)+(a+3d)=360°
⇒4a=360°
⇒a=90°
Now common difference=10°
⇒(Second angle)-(First angle)=10°
⇒(a-d)-(a-3d)=10°
⇒a-d-a+3d=10°
⇒2d=10°
⇒d=5°
So when a=90° and d=5° –
a-3d=75°
a-d=85°
a+d=95°
a+3d=105°
So required angles are 75°,85°,95° and 105°.
问题 10:将 207 拆分为三个部分,使得它们在 AP 中,两个较小部分的乘积是 4623。
解决方案:
Let the first term of the A.P. is=a
and common difference is=d
let three parts are=(a-d),a,(a+d)
Since (a-d),a and (a+d) are parts of 207—
⇒(a-d)+a+(a+d)=207
⇒3a=207
⇒a=69
Now according to given condition-
⇒(a-d)*a=4623
⇒(69-d)*69=4623
dividing by 69-
⇒69-d=67
⇒d=2
So when a=69 and d=2 –
a-d=67
a=69
a+d=71
means required three parts of 207 are 67,69 and 71.