NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3.
- Sequences and Series Class 11 Ex 9.1
- Sequences and Series Class 11 Ex 9.2
- Sequences and Series Class 11 Ex 9.4
|Chapter Name||Sequences and Series|
|Number of Questions Solved||32|
NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3
Find the 20th and nth terms of the G.P. , ….
Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
We have, as = 192, r = 2
The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.
We are given
The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7th term.
We have a= -3, a4 = (a2)2
Which term of the following sequences:
For what values of x, the numbers , x, are in G.P.?
Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:
0.14, 0.015, 0.0015, …. 20 items.
In the given G.P.
, …. n terms
In the given G.P.
1, -a, a2,- a3 … n terms (if a ≠ -1)
In the given G.P.. a = 1, r = -a
x3, x5, 7, ….. n terms (if ≠±1).
In the given G.P., a = x3, r = x2
The sum of first three terms of a G.P. is and 10 their product is 1. Find the common ratio and the terms.
Let the first three terms of G.P. be ,a,ar, where a is the first term and r is the common ratio.
How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?
Let n be the number of terms we needed. Here a = 3, r = 3, Sn = 120
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 a1, a2, a3, a4, a5, a6 be the first six terms of the G.P.
Given a G.P. with a = 729 and 7th term 64, determine S7.
Let a be the first term and the common ratio be r.
Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.
Let a1 a2 be first two terms and a3 a5 be third and fifth terms respectively.
According to question
If the 4th, 10th and 16th terms of a G.P. are x, y and z respectively. Prove that x, y, z are in G.P.
Let a be the first term and r be the common ratio, then according to question
Find the sum to n terms of the sequence, 8, 88, 888, 8888 ………
This is not a G.P., however we can relate it to a G.P. by writing the terms as Sn= 8 +88 + 888 + 8888 + to n terms
Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2,
On multiplying the corresponding terms of sequences, we get 256, 128, 64, 32 and 16, which forms a G.P. of 5 terms
Show that the products of the corresponding terms of the sequences a, ar, ar2, ………… arn-1 and A, AR, AR2, …….. , ARn-1 form a G.P., and find the common ratio.
On multiplying the corresponding terms, we get aA, aArR, aAr2R2,…… aArn-1Rn-1. We can see that this new sequence is G.P. with first term aA & the common ratio rR.
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 4th by 18.
Let the four numbers forming a G.P. be a, ar, ar2, ar3
According to question,
If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that aq-r br-p cp-q = 1.
Let A be the first term and R be the common ratio, then according to question
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 P2 – (ab)n.
Let r be the common ratio of the given G.P., then b = nth term = arn-1
Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is
Let the G.P. be a, ar, ar2, ……
Sum of first n terms = a + ar + ……. + arn-1
If a, b,c and d are in G.P., show that
(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2
We have a, b, c, d are in G.P.
Let r be a common ratio, then
Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Let G1, G2 be two numbers between 3 and 81 such that 3, G1 G2,81 is a G.P.
Find the value of n so that may be the geometric mean between a and b.
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio .
Let a and b be the two numbers such that a + b = 6
If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are .
Let a and b be the numbers such that A, G are A.M. and G.M. respectively between them.
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?
There were 30 bacteria present in the culture originally and it doubles every hour. So, the number of bacteria at the end of successive hours form the G.P. i.e., 30, 60, 120, 240, …….
What will Rs. 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
We have, Principal value = Rs. 500 Interest rate = 10% annually
If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Let α & β be the roots of a quadratic equation such that A.M. & G.M. of α, β are 8 and 5 respectively.
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