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Class 11 RD Sharma Solutions – Chapter 20 Geometric Progressions- Exercise 20.1 | Set 2

  • Last Updated : 28 Dec, 2020

Question 11. If the GP’s 5, 10, 20,…….. and 1280, 640, 320,……..have their nth terms equal, find the value of n.

Solution: 

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*rn-1 = a1’*r’n-1

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

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

22n-2 = 256

22n-2 = (2)8

2n-2 = 8

2n = 10 

n = 5

Question 12. If the 5th, 8th and 11th terms of a GP are p, q and s respectively. Prove that q2 = ps.

Solution:

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.

Question 13. The 4th term of a GP is square of its 2nd term and the first term is -3. Find its 7th term.

Solution:

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  

Question 14. In GP the 3rd term is  24 and the 6th term is 192. Find the 10th term.

Solution:

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.

Question 15. If a, b, c, d and p are different real numbers such that:

(a2+b2+c2)p2 – 2(ab+bc+cd)p + (b2+c2+d2) ≤ 0, then show that a, b, c and d are in GP.

Solution:

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) ≤ 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.

Question 16. If (a+bx)/(a-bx) = (b+cx)/(b-cx) = (c+dx)/(c-dx) (x≠0), then show that a, b, c and d are in GP.

Solution:

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.

Question 17. If the pth and qth terms of a GP are q and p respectively. Show that (p+q)th term is (qp/pq)1/p-q.

Solution:

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*rq-1)/(a1*rp-1) = p/q

r(q-p) = p/q

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

ap = q

a1*rp-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)]

            = [qp/pq] 1/(p-q)

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.

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