Chapter 9 Sequences and Series Ex 9.3 Solutions
Question - 21 : - Find four numbers forming a geometric progression in which third term isgreater than the first term by 9, and the second term is greater than the 4th by18.
Answer - 21 : -
Consider a tobe the first term and r to be the common ratioof the G.P.
Then,
a1 = a, a2 = ar, a3 = ar2, a4 = ar3
From the question, wehave
a3 = a1 + 9
ar2 = a + 9 … (i)
a2 = a4 + 18
ar = ar3 + 18 … (ii)
So, from (1) and (2),we get
a(r2 – 1) = 9 … (iii)
ar (1– r2) = 18 … (iv)
Now, dividing (4) by(3), we get
-r = 2
r = -2
On substituting thevalue of r in (i), we get
4a = a +9
3a = 9
∴ a =3
Therefore, the firstfour numbers of the G.P. are 3, 3(– 2), 3(–2)2, and 3(–2)3
i.e., 3¸–6, 12, and–24.
Question - 22 : - If the pth, qth and rth termsof a G.P. are a, b and c, respectively. Prove thataq-r br-p cp-q = 1
Answer - 22 : -
Let’s take A tobe the first term and R to be the common ratio of the G.P.
Then according to thequestion, we have
ARp–1 = a
ARq–1 = b
ARr–1 = c
Then,
aq–r br–p cp–q
= Aq–r × R(p–1)(q–r) × Ar–p × R(q–1)(r–p) × Ap–q × R(r –1)(p–q)
= Aq – r + r – p + p – q × R (pr – pr – q + r)+ (rq – r + p – pq) +(pr – p – qr + q)
= A0 × R0
= 1
Hence proved.
Question - 23 : - If the first and the nth term of a G.P.are a ad b, respectively, and if P isthe product of n terms, prove that P2 =(ab)n.
Answer - 23 : -
Given, the first termof the G.P is a and the last term is b.
Thus,
The G.P. is a, ar, ar2, ar3,… arn–1, where r is thecommon ratio.
Then,
b = arn–1 … (1)
P = Product of n terms
= (a) (ar)(ar2) … (arn–1)
= (a × a ×…a)(r × r2 × …rn–1)
= an r 1+ 2 +…(n–1) … (2)
Here, 1, 2, …(n –1) is an A.P.
And, the product of nterms P is given by,
Question - 24 : - Show that the ratio of the sum of first n termsof a G.P. to the sum of terms from 
Answer - 24 : -
Let a bethe first term and r be the common ratio of the G.P.
Since there are n termsfrom (n +1)th to (2n)th term,
Sum of terms from(n +1)th to (2n)th term
a n +1 = ar n+ 1 – 1 = arn
Thus, required ratio =
Thus, the ratio of thesum of first n terms of a G.P. to the sum of terms from (n +1)th to (2n)th term is 
Answer - 25 : -
Given, a, b, c, d arein G.P.
So, we have
bc = ad … (1)
b2 = ac … (2)
c2 = bd … (3)
Taking the R.H.S. wehave
R.H.S.
= (ab + bc + cd)2
= (ab + ad + cd)2 [Using (1)]
= [ab + d (a + c)]2
= a2b2 +2abd (a + c) + d2 (a + c)2
= a2b2 +2a2bd +2acbd + d2(a2 + 2ac + c2)
= a2b2 +2a2c2 + 2b2c2 + d2a2 +2d2b2 + d2c2 [Using (1) and (2)]
= a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2
= a2b2 + a2c2 + a2d2 + b2 × b2 + b2c2 + b2d2 + c2b2 + c2 × c2 + c2d2
[Using(2) and (3) and rearranging terms]
= a2(b2 + c2 + d2)+ b2 (b2 + c2 + d2)+ c2 (b2+ c2 + d2)
= (a2 + b2 + c2)(b2 + c2 + d2)
= L.H.S.
Thus, L.H.S. = R.H.S.
Therefore,
(a2 +b2 + c2)(b2 + c2 + d2)= (ab + bc + cd)2
Question - 26 : - Insert two numbers between 3 and 81 so that the resulting sequence isG.P.
Answer - 26 : -
Let’s assume G1 and G2 tobe two numbers between 3 and 81 such that the series 3, G1, G2,81 forms a G.P.
And let a bethe first term and r be the common ratio of the G.P.
Now, we have the 1st termas 3 and the 4th term as 81.
81 = (3) (r)3
r3 = 27
∴ r =3 (Taking real roots only)
For r =3,
G1 = ar = (3) (3) = 9
G2 = ar2 = (3) (3)2 =27
Therefore, the twonumbers which can be inserted between 3 and 81 so that the resulting sequencebecomes a G.P are 9 and 27.
Question - 27 : - Find the value of n so that 
may be the geometricmean between a and b.
Answer - 27 : -
We know that,
The G. M. of a and b is givenby √ab.
Then from thequestion, we have
By squaring bothsides, we get
Question - 28 : - The sum of two numbers is 6 times theirgeometric mean, show that numbers are in the ratio
Answer - 28 : -
Consider the twonumbers be a and b.
Then, G.M. = √ab.
From the question, wehave
Question - 29 : - If A and G be A.M. and G.M., respectively between two positive numbers, prove that the
numbers are
.
Answer - 29 : -
Given that A and G areA.M. and G.M. between two positive numbers.
And, let these twopositive numbers be a and b.
Question - 30 : - The number of bacteria in a certain culture doubles every hour. If therewere 30 bacteria present in the culture originally, how many bacteria will bepresent at the end of 2nd hour, 4th hourand nth hour?
Answer - 30 : -
Given, the number ofbacteria doubles every hour. Hence, the number of bacteria after every hourwill form a G.P.
Here we have, a =30 and r = 2
So, a3 = ar2 =(30) (2)2 = 120
Thus, the number ofbacteria at the end of 2nd hour will be 120.
And, a5 = ar4 =(30) (2)4 = 480
The number of bacteriaat the end of 4th hour will be 480.
an +1 = arn = (30) 2n
Therefore, the numberof bacteria at the end of nth hour will be 30(2)n.