NCERT Solutions class-11 Maths Exercise 9.3 Part-2

myCBSEguide App

myCBSEguide App

CBSE, NCERT, JEE Main, NEET-UG, NDA, Exam Papers, Question Bank, NCERT Solutions, Exemplars, Revision Notes, Free Videos, MCQ Tests & more.

Install Now

 

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

Ans. Let be the first term and be the common ratio of given G.P.

Given:

……..(i)

And

Putting in eq. (i), we get

Therefore, required G.P. is

Putting in eq. (i), we get

Therefore, required G.P. is


17. If the 4th, 10th and 16th terms of a G.P. are and respectively. Prove that are in G.P.

Ans. Let be the first term and be the common ratio of given G.P.

……….(i)

……….(i)

……….(i)

From eq. (ii),

[From eq. (i) and (iii)]

are in G.P.


18. Find the sum to terms of the sequences 8, 88, 888, 8888, ……

Ans. Here

=

=


19. Find the sum of the product of the corresponding terms of the sequences 2, 4, 16, 32 and 128, 32, 8, 2,

Ans. Multiplying the corresponding terms of the given sequences 2, 4, 16, 32 and 128, 32, 8, 2,

are in G.P.

Here and

when

=

= 496


20. Show that the products of the corresponding terms of the sequences and form a G.P. and find the common ratio.

Ans. Multiplying the corresponding terms of the given sequences, we have

are in G.P.

Here First term = and common ratio =


21. 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 by 4th by 18.

Ans. Let the four numbers in G.P. be

and

Now,

……….(i)

And

……….(ii)

Dividing eq. (ii) by eq. (i), we have

Putting value of in eq. (i), we get

Therefore, the required numbers are


22. If the and terms of a G.P. are and respectively. Prove that

Ans. Let A be the first term and R be the common ratio of given G.P.

……….(i)

……….(ii)

……….(iii)

Now, L.H.S. =

=

=

= = = 1 = R.H.S.


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

Ans. Here, first term of G.P. is

And

……….(i)

Given: P =

P =

[Squaring both sides]

[From eq. (i)]


24. Show that the ratio of the sum of first terms of a G.P. to the sum of terms from to term is

Ans. Let be the first term and be the common ratio of given G.P.

Then

=


25. If and are in G.P., show that

Ans. Let be the common ratio of given G.P.

Then and

Now, L.H.S. =

=

= =

R.H.S. =

=

=

= =

Therefore, L.H.S. = R.H.S.


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

Ans. Let be two numbers between 3 and 81 such that are in G.P.

Here and

And

Therefore, the required numbers are 9 and 27.


27. Find the value of so that may be the geometric mean between and .

Ans. Since, G.M. between two numbers and is

According to question,


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

Ans. Given:

Applying component and dividend, we get

Again, Applying component and dividend, we get

Squaring both sides,

Therefore, the numbers are in the ratio


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

Ans. Given: A = and G =

Now,

=

=

=

= =

= and

= and

= and


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 hour?

Ans. Bacteria present in the culture originally = 30

Since the bacteria doubles itself after each hour, then the sequence of bacteria after each hour is a G.P.

Here and

Bacteria at the end of 2nd hour =

And Bacteria at the end of 4th hour =

And Bacteria at the end of hour =


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

Ans. Original amount = Rs. 500, Rate of interest = 10% compounded annually

Interest of one year = = Rs. 50

And Amount after one year = 500 + 50 = Rs. 550

Here and

Therefore, amount after 10 years =


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

Ans. Let and be the roots of required quadratic equation.

Then A.M. =

And G.M. =

Now, Quadratic equation

Therefore, required equation is .


http://mycbseguide.com/examin8/




4 thoughts on “NCERT Solutions class-11 Maths Exercise 9.3 Part-2”

Leave a Comment