**32.** If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation

Let roots of the quadratic equation be a and b.
According to given condition,
We know that
Thus, the quadratic equation =

**31.** What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

Given: Bank pays an annual interest rate of 10% compounded annually.
Rs 500 amounts are deposited in the bank.
At the end of the first year, the amount
At the end of the second year, the amount
At the end of the third year, the amount
At the end of 10 years, the amount
Thus, at the end of...

**30. **The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour ?

The number of bacteria in a certain culture doubles every hour.It forms GP.
Given : a=30 and r=2.
Thus, bacteria present at the end of the 2nd hour, 4th hour and nth hour are 120,480 and respectively.

**29. ** If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are

If A and G be A.M. and G.M., respectively between two positive numbers,
Two numbers be a and b.
...................................................................1
...........................................................................2
We know
Put values from equation 1 and 2,
..................................................................3
From 1 and 3 , we have
Put value...

**28.** The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio

Let there be two numbers a and b
geometric mean
According to the given condition,
.............................................................(1)
Also,
.......................................................(2)
From (1) and (2), we get
Putting the value of 'a' in (1),
Thus, the ratio is

**27.** Find the value of n so that may be the geometric mean between a and b.

**26.** Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Let A, B be two numbers between 3 and 81 such that series 3, A, B,81 forms a GP.
Let a=first term and common ratio =r.
For ,
The, required numbers are 9,27.

**25**. If a, b, c and d are in G.P. show that

If a, b, c and d are in G.P.
To prove :
RHS :
Using equation (1) and (2),
= LHS
Hence proved

**24.** Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from

term is

.

Let first term =a and common ratio = r.
Since there are n terms from (n+1) to 2n term.
Sum of terms from (n+1) to 2n.
Thus, the required ratio =
Thus, the common ratio of the sum of first n terms of a G.P. to the sum of terms from term is .

**23. ** If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that

.

Given : First term =a and n th term = b.
Common ratio = r.
To prove :
Then ,
P = product of n terms
Here, is a AP.
Put in equation (2),
Hence proved .

**22.** If the terms of a G.P. are a, b and c, respectively. Prove that

To prove :
Let A be the first term and R be common ratio.
According to the given information, we have
L.H.S :
=RHS
Thus, LHS = RHS.
Hence proved.

**2.1** Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the by 18.

Let first term be a and common ratio be r.
Given : the third term is greater than the first term by 9, and the second term is greater than the by 18.
Dividing equation 2 by 1 , we get
Putting value of r , we get
Thus, four terms of GP are

**20. ** Show that the products of the corresponding terms of the sequences

form a G.P, and find the common ratio.

**19. **Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2,1/2

**18.** Find the sum to n terms of the sequence, 8, 88, 888, 8888… .

**17. ** If the terms of a G.P. are x, y and z, respectively. Prove that x,y, z are in G.P.

Let x,y, z are in G.P.
Let first term=a and common ratio = r
Dividing equation 2 by 1, we have
Dividing equation 3 by 2, we have
Equating values of , we have
Thus, x,y,z are in GP

**16. ** Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term

Given : sum of the first two terms is – 4 and the fifth term is 4 times the third term
Let first term be a and common ratio be r
If r=2, then
If r= - 2, then
Thus, required GP is or

**15. ** Given a G.P. with a = 729 and term 64, determine

**14. **The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

Let GP be
Given : The sum of first three terms of a G.P. is 16
Given : the sum of the next three terms is128.
Dividing equation (2) by (1), we have
Putting value of r =2 in equation 1,we have

**13.** How many terms of G.P. , … are needed to give the sum 120?

G.P.= , …............
Sum =120
These terms are GP with a=3 and r=3.
Hence, we have value of n as 4 to get sum of 120.

- Previous
- Next

Exams

Articles

Questions