Question 31
Question: In a finite G.P., the product of the terms equidistant from the beginning and the end is always same and equal to the product of the first and last term.
Answer: Let a1, a2, a3,- - - -,an be a finite GP with common ratio = r.
Then ak = kth term from the beginning = ark-1
And an-k+1 is the kth term from the other end.
Hence the problem.
Question 32
Question: The first term of a GP is 1. The sum of the third and fifth terms is 90. Find the common ratio of the GP.
Answer: 
Let r = common ratio
The common ratio is either +3 or -3.
Question 33
Question: For each geometric progression, find the common ratio. Then tell whether the ratio is a growth factor or decay factor or neither. Find also the nth term.
ii. 1, 1.1, 1.21, 1.331,....
iii. -1, 1, -1, 1, -1,....
Answer: 
Since r = 1.1 > 1, r is a growth factor.
Here the progression is oscillating.
Question 34
Question: Use the given values for a and r to find the first four terms of a GP and the nth term.
1. a = 2, r = 5
2. a = -2, r =3
4. a = x, r = y
Answer: 
Question 35
Question: What is the decay factor for the value of a car that depreciates 20% per year. Give the value at the end of 0, 1, 2, 3 and 4 years for a car that costs Rs. 300000/- as a sequence.
Answer: The value of car at the 0th year = Rs. 300000
The value of car at the end of 1st year
= (100 - 20)% of Rs. 300000 = Rs. 240000
The value of car at the end of 2nd year = Rs. 192000
The value of car at the end of 3rd year = Rs. 153600
The value of car at the end of 4th year = Rs. 122880
Question 36
Question: Find three numbers in GP such that their sum is 38 and their product is 1728.
Answer: 
The numbers are
Question 37
Question: Find four numbers in GP in which the third term is greater than the first by 9 and the second term is greater than the fourth by 18.
Answer: Let the four numbers in GP be a, ar, ar2, ar3
Third term-first term = 9
ar2 - a = 9... (i)
Second term-fourth term = 18
ar - ar3 = 18... (ii)
Dividing (ii) by (i) we get
r = -2... (iii)
Now putting r = -2 in (i)
The numbers are
Question 38
Question: If a, b, c, d are in GP, Prove that a+b, b+c, c+d are also in GP.
Answer: From the problem we have a, b, c, d are in GP
Now a+b, b+c, c+d will be in GP if,
Question 39
Question: If a, b, c are in GP, prove that the following are also in GP.
Answer: a, b, c, d are in GP
Question 40
Question: If a, b, c are three consecutive terms of an AP, then prove that ka, kb and kc are three consecutive terms in GP, where k is a non zero real number.
Answer: Since a, b, c are in AP
b - a = c - b...(i)
If ka, kb, kcwill be in G.P.,
then kb-a = kc-b. This is true because of (i).
Hence ka, kb and kc are in GP.
