11类RD Sharma解决方案–第20章几何级数-练习20.1 |套装2

问题11。如果GP的5、10、20，……和1280、640、320，……具有第n个项，则求n的值。

解决方案：

For GP 5, 10, 20,…………..

a1 = 5,

r = a2/a1 =10/5 = 2

For GP 1280, 640, 320,…………..

a1′ = 1280,

r’ = a2’/a1′ =640/1280 = 1/2

Given nth terms are equal 

Hence, a1*r^n-1 = a1’*r’^n-1

5*2^n-1 = 1280*(1/2)^n-1

2^n-1 * 2^n-1 = 1280/5

2^2n-2 = 256

2^2n-2 = (2)⁸

2n-2 = 8

2n = 10 

n = 5

问题12.如果GP的第5，第8和第11项分别是p，q和s。证明q ² = ps。

解决方案：

Let a1 be first term and common ratio be r of given GP 

Given a5 = p, a8 = q, a11 = s

a8/a5 = q/p

(a1*r⁷)/(a1*r⁴) = q/p

r³ = q/p –Equation I

a11/a8 = s/q

(a1*r¹⁰)/(a1*r⁷) = s/q

r³ = s/q — Equation II

From Equation, I and Equation II

q/p = s/q

q² = ps

Hence, proved that q² = ps.

问题13. GP的第四项是其第二项的平方，而第一项是-3。找到它的第七个学期。

解决方案：

Let a1 be first term and common ratio be r of given GP 

Given a4 = (a2)², a1 = -3

a1*r3 = (a1*r)²

-3*r3 = (-3*r)²

-3*r³ = 9*r²

r = -3

7th term of given GP is given by

a7 = a1*r6

     = -3*(-3)⁶

     = (-3)⁷

     = -2187  

问题14.在GP中，第三学期是24，第六学期是192。找到第十学期。

解决方案：

Let a1 be first term and common ratio be r of given GP 

Given a3 = 24, a6 = 192

a6/a3 = 8

(a1*r⁵)/(a1*r²) = 8

r³ = 8

r = 2

We have a3=24

a1*r² = 24

a1*2² = 24

a1*4 = 24

a1 = 6

The 10th term of given GP is given by

a10 = a1*r⁹

         = 6*2⁹

       = 3072.

问题15：如果a，b，c，d和p是不同的实数，则：

(a ² + b ² + c ² )p ² – 2(ab + bc + cd)p +(b ² + c ² + d ² )≤0，则表明a，b，c和d在GP中。

解决方案：

Given, (a²+b²+c²)p² – 2(ab+bc+cd)p + (b²+c²+d²) ≤ 0

(a²p²+b²p²+c²p²) – 2(abp+bcp+cdp) + (b²+c²+d²) ≤ 0

(a²p²+b²-2abp) +  (b²p²+c²-2bcp) +  (c²p²+d²-2cdp)  ≤ 0

(ap-b)² + (bp-c)² + (cp-d)²  ≤ 0

Sum of squares cannot be less than 0.

Hence, (ap-b)² + (bp-c)² + (cp-d)²  = 0

(ap-b)² = 0

p = b/a

(bp-c)² = 0

p = c/b

(cp-d)² =0

p = d/c

b/a = c/b = d/c

Hence, a, b, c and d are in GP.

问题16.如果(a + bx)/(a-bx)=(b + cx)/(b-cx)=(c + dx)/(c-dx)(x≠0)，则表明a， b，c和d在GP中。

解决方案：

Given (a+bx)/(a-bx) = (b+cx)/(b-cx) 

Applying componendo and dividendo 

(a+bx)+(a-bx) / (a+bx)-(a-bx) = (b+cx)+(b-cx) / (b+cx)-(b-cx)

2a/2bx = 2b/2cx

a/b =b/c –Equation I

Similarly, (b+cx)/(b-cx) = (c+dx)/(c-dx)

Applying componendo and dividendo

(b+cx)+(b-cx) / (b+cx)-(b-cx) = (c+dx)+(c-dx) / (c+dx)-(c-dx)

2b/2cx = 2c/2dx

b/c =c/d –Equation II

From Equation, I and Equation II 

a/b = b/c = c/d

b/a = c/b = d/c

Hence, a, b, c and d are in GP.

问题17.如果GP的第p个和第q个项分别是q和p。证明第(p + q)个项是(qp / pq)1 / pq。

解决方案：

Let the first term of given GP be a1 and common ratio be r.

Given, ap = q and aq = p

aq/ap = p/q

(a1*r^q-1)/(a1*r^p-1) = p/q

r^(q-p) = p/q

r = (p/q)^1/(q-p)

ap = q

a1*r^p-1 = q

a1*(p/q)^(p-1)/(q-p) = q

a1 = ((q/p)^(p-1)/(q-p) )* q

The (p+q)th term of given GP is given by 

a(p+q) = a1*r^(p+q-1)

            = [((q/p)^(p-1)/(q-p) )* q]*[(p/q)^1/(q-p)]^(p+q-1)

            = [((q/p)^(p-1)/(q-p) )* q]*[(p/q)^{(p+q-1)/(q-p)}]

            = q*[(q/p)^(p-1)/(q-p)]*[(p/q)^{(p+q-1)/(q-p)}]

            = q*[(p/q)^{((1-p)+(p+q-1))/(q-p)} ]

            = q*[(p/q)^q/(q-p)]

            = q*[(q/p)^q/(p-q)]

            = [q^p/p^q] ^1/(p-q)