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All questions of Sequences & Series of Real Numbers for Mathematics Exam
If 
then
a)| a | ≥ 1 b)| a | ≤ 1c)| a | > 1d)| a | < 1Correct answer is option 'D'. Can you explain this answer?
|
Chirag Verma answered |
Correct Answer :- D
Explanation : For |a|<1
b = 1/a, |b|>1
lim (n→∞) an = lim (n→∞) (1/b)n
= lim(n→∞) 1/bn
= 1/(±∞)
= 0
A convergent sequence is a Cauchy sequence, if it is a- a)sequence of real numbers
- b)sequence of rational numbers
- c)sequence of irrational numbers
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
A convergent sequence is a Cauchy sequence, if it is a
a)
sequence of real numbers
b)
sequence of rational numbers
c)
sequence of irrational numbers
d)
None of the above
|
|
Chirag Verma answered |
ANSWER :- b
Solution :- If {an}∞n=1 is a cauchy sequence of real numbers and if there is a sub-sequence of this sequence, {anj}∞j=1 which converges to a real number L, then I need to show that the sequence {an}∞n=1 converges to the real number L.
The sequence {xn}, where xn = nl/n, converge to- a)1
- b)0
- c)1/2
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The sequence {xn}, where xn = nl/n, converge to
a)
1
b)
0
c)
1/2
d)
None of these
|
Om Pavan Koushik answered |
One can view the sequence as a function and take log on both sides and evaluate using l'hospital rule.
The sequence <sn> = 
is - a)not monotonic but bounded
- b)not monotonic and not bounded
- c)monotonic but not bounded
- d)monotonic and bounded
Correct answer is option 'A'. Can you explain this answer?
The sequence <sn> = is
a)
not monotonic but bounded
b)
not monotonic and not bounded
c)
monotonic but not bounded
d)
monotonic and bounded
|
|
Chirag Verma answered |
Here lower bound = l = − 1/2 , & upper bounded u = 1. But the given sequence is neither increasing nor decreasing. Thus, <sn> is bounded but not monotonic.
convergent, if
Let San be a convergent series of positive terms and let Sbn be a divergent series of positive terms. Then,
- a)the sequence < an > is convergent and < bn > is not convergent
- b)the sequence < an > converges to 0
- c)the sequence < bn > does not converge to 0
- d)the sequence < bn > diverges to ∞
Correct answer is option 'B'. Can you explain this answer?
Let San be a convergent series of positive terms and let Sbn be a divergent series of positive terms. Then,
a)
the sequence < an > is convergent and < bn > is not convergent
b)
the sequence < an > converges to 0
c)
the sequence < bn > does not converge to 0
d)
the sequence < bn > diverges to ∞
|
Chhotu Yadav answered |
Answer is abc
is equal to- a)0
- b)1
- c)e
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
is equal toa)
0
b)
1
c)
e
d)
None of these
|
Ibrahim Parekh answered |
Take log on both the sides.
If Sn denoted the sum of n term of the series 
- a)Sn > n2
- b)Sn > n
- c)
- d)Sn = ∞
Correct answer is option 'C'. Can you explain this answer?
If Sn denoted the sum of n term of the series
a)
Sn > n2
b)
Sn > n
c)
d)
Sn = ∞
|
|
Chirag Verma answered |
If Sn denote the sum of n terms of the series 
then Sn will be > 
Explanation:
A: Sn > n^2
- The given comparison tells us that the sum of the first n terms of the series is greater than n^2.
- This could be true for certain series where the sum of terms grows faster than n^2, such as series with exponential growth.
- However, this comparison alone does not provide enough information to determine the behavior of the series for all values of n.
B: Sn > n
- This comparison indicates that the sum of the first n terms of the series is greater than n itself.
- Similar to the first comparison, this could be true for certain series with increasing terms.
- Again, this comparison alone does not give a definitive answer about the convergence or divergence of the series.
C: Sn = ∞
- This comparison suggests that the sum of the terms of the series diverges to infinity.
- If the sum of the series approaches infinity as n increases, it indicates that the series does not converge.
- This comparison implies that the series may be divergent and does not have a finite sum.
D: Sn = ∞
- This comparison reiterates the previous point that the sum of the series diverges to infinity.
- It reinforces the idea that the series does not have a finite sum and diverges as n increases.
Conclusion:
- Based on the comparisons provided, it is most likely that the series described does not converge and has an infinite sum. This is indicated by the comparisons stating that the sum of the terms exceeds n^2 and n, and ultimately diverges to infinity.
then Sn will be > Explanation:A: Sn > n^2
- The given comparison tells us that the sum of the first n terms of the series is greater than n^2.
- This could be true for certain series where the sum of terms grows faster than n^2, such as series with exponential growth.
- However, this comparison alone does not provide enough information to determine the behavior of the series for all values of n.
B: Sn > n
- This comparison indicates that the sum of the first n terms of the series is greater than n itself.
- Similar to the first comparison, this could be true for certain series with increasing terms.
- Again, this comparison alone does not give a definitive answer about the convergence or divergence of the series.
C: Sn = ∞
- This comparison suggests that the sum of the terms of the series diverges to infinity.
- If the sum of the series approaches infinity as n increases, it indicates that the series does not converge.
- This comparison implies that the series may be divergent and does not have a finite sum.
D: Sn = ∞
- This comparison reiterates the previous point that the sum of the series diverges to infinity.
- It reinforces the idea that the series does not have a finite sum and diverges as n increases.
Conclusion:
- Based on the comparisons provided, it is most likely that the series described does not converge and has an infinite sum. This is indicated by the comparisons stating that the sum of the terms exceeds n^2 and n, and ultimately diverges to infinity.
The series 2 + 4 + 6 + 8 + ...is
- a)divergent
- b)convergent
- c)Both of these
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The series 2 + 4 + 6 + 8 + ...is
a)
divergent
b)
convergent
c)
Both of these
d)
None of these
|
|
Chirag Verma answered |
The sequence of n th partial sum of given series is unbounded and goes to infinity.
since sequence of n th partial sum is unbounded, it is not convergent therefore the given series
is not convergent i.e. divergent.
For the sequence 1, 7, 25, 79, 241, 727 … simple formula for {an} is ____________ - a)3n+1 – 2
- b)(-3)n + 4
- c)n2 – 2
- d)3n – 2
Correct answer is option 'D'. Can you explain this answer?
For the sequence 1, 7, 25, 79, 241, 727 … simple formula for {an} is ____________
a)
3n+1 – 2
b)
(-3)n + 4
c)
n2 – 2
d)
3n – 2
|
|
Chirag Verma answered |
The ratio of consecutive numbers is close to 3. Comparing these terms with the sequence of {3n} which is 3, 9, 27 …. Comparing these terms with the corresponding terms of sequence {3n} and the nth term is 2 less than the corresponding power of 3.
then the value of 
Let < an> —> a. Let for every positive integer k, Ak be the set of all positive integer n such that | an- a | < —1/k. Then,- a)Ak is finite for all k
- b)Ak is finite for some k
- c)every Ak contains all but finitely many positive integers
- d)Ak contains all positive integers
Correct answer is option 'D'. Can you explain this answer?
Let < an> —> a. Let for every positive integer k, Ak be the set of all positive integer n such that | an- a | < —1/k. Then,
a)
Ak is finite for all k
b)
Ak is finite for some k
c)
every Ak contains all but finitely many positive integers
d)
Ak contains all positive integers
|
Tapas Bhowmik answered |
I think C is correct
then which one of the following statement is correct?- a)Both S1 and S2 are convergent
- b)S1 is divergent and S2 is convergent
- c)S1 is convergent and S2 is divergent
- d)Both S1 and S2 are divergent
Correct answer is option 'B'. Can you explain this answer?
then which one of the following statement is correct?a)
Both S1 and S2 are convergent
b)
S1 is divergent and S2 is convergent
c)
S1 is convergent and S2 is divergent
d)
Both S1 and S2 are divergent
|
Veda Institute answered |
So, we can see that from S1 and S2 that is divergent and S2 is convergent.
Match list I with list II and select the correct answer
- a)a
- b)b
- c)c
- d)d
Correct answer is option 'C'. Can you explain this answer?
Match list I with list II and select the correct answer
a)
a
b)
b
c)
c
d)
d
|
|
Chirag Verma answered |
is also convergent series.
∴ un is also convergent series.
Hence, by d’Alembert test
implies un is divergent series.
By Cauchy condensation test, ∑μn is convergent series.
(D)
By Leibnitz’s test, the series is convergent.
Also,
is divergent series. So, the given series is converges conditionally.
- a)is not bounded
- b)may or may not be bounded
- c)convergent
- d)may or may not be convergent
Correct answer is option 'A'. Can you explain this answer?
a)
is not bounded
b)
may or may not be bounded
c)
convergent
d)
may or may not be convergent
|
Kartikey Mehrotra answered |
Simply by cauchy second theorem on limits if lim Sn = l, then lim Sn/n= l also.
Therefore here Sn= l and is convergent to L. option c is correct
Therefore here Sn= l and is convergent to L. option c is correct
The series 
- a)conditional convergent
- b)absolutely convergent
- c)divergent
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
The series
a)
conditional convergent
b)
absolutely convergent
c)
divergent
d)
None of the above
|
|
Chirag Verma answered |
Let the given series be denoted by ∑μn, then
∑|μn|
Compare this series ∑vn with the auxiliary series,
Then,
which is finite quantity.
Hence, ∑un and ∑wn are either both convergent or both divergent. But ∑wn
Hence, the series ∑vn is divergent.
Also, in the series ∑un, we find that its term are alternately positive and negative, its terms are continuously decreasing and
Thus, all condition of Leibnitz’s test are satisfied and as such ∑un is convergent. H ence, the given series ∑un is conditionally convergent.
∑|μn|
Compare this series ∑vn with the auxiliary series,
Then,
which is finite quantity.
Hence, ∑un and ∑wn are either both convergent or both divergent. But ∑wn
Hence, the series ∑vn is divergent.
Also, in the series ∑un, we find that its term are alternately positive and negative, its terms are continuously decreasing and
Thus, all condition of Leibnitz’s test are satisfied and as such ∑un is convergent. H ence, the given series ∑un is conditionally convergent.
The sequence 
- a)divergent
- b)convergent to 0
- c)convergent to 1
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The sequence
a)
divergent
b)
convergent to 0
c)
convergent to 1
d)
None of these
|
|
Chirag Verma answered |
Correct Answer :- b
Explanation : Suppose otherwise, that there exists a number L implies R and a positive integer N such that
| f(n) - L | < e {for all } e > 0 {for all } n > N.
Since N is a positive integer, we know 4N > N and 4N+2 > N.
But,
f(4N) = cos (2N pi) = 1 and f(4N+2) = cos((2N+1)pi) = -1
Taking e = 1/2
= |1 - L| < 1/2 and |-1-L| < 1/2
=> |1+L| < 1/2
But, these imply
|1-L| + |1+L| < 1.
By the triangle inequality we then have
|1 - L + 1 + L | < 1
=> 2 < 1
a contradiction. Hence, there is no such limit L.
Therefore, the sequence converges to zero.
For what value of x the infinite series 
diverges?- a)x < e
- b)x > e
- c)
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
For what value of x the infinite series diverges?
diverges?a)
x < e
b)
x > e
c)
d)
None of these
|
|
Chirag Verma answered |
If e/x > 1 i.e.., x < e, then the given series diverges.
If {xn} and {yn} are two convergent sequence such that xn < yn, 
n ∈ N, then- a)
- b)
- c)
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
If {xn} and {yn} are two convergent sequence such that xn < yn, n ∈ N, then
n ∈ N, thena)
b)
c)
d)
None of these
|
|
Chirag Verma answered |
The correct option is Option A.
Let lim xn = x and lim yn = y and zn = yn - xn
n->∞ n->∞
Then (zn) is a convergent sequence such that zn > 0 ∇ n ∊N
and lim zn = y - x.
n->∞
Now (zn) is a convergent sequence of real numbers and zn > 0 ∇ n ∊N
So, lim zn ≥ 0
n->∞
So, y - x ≥ 0
=> y ≥ x
=> lim yn ≥ lim xn
n->∞ n->∞
Hence, proved.
If a sequence is not a Cauchy sequence, then it is a
- a)divergent sequence
- b)convergent sequence
- c)bounded sequence
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
If a sequence is not a Cauchy sequence, then it is a
a)
divergent sequence
b)
convergent sequence
c)
bounded sequence
d)
None of the above
|
Dronacharya Institute answered |
A sequence that is not a Cauchy sequence is called a divergent sequence. A Cauchy sequence is a sequence whose elements get closer and closer together as the sequence progresses.
The series
- a)convergent
- b)divergent
- c)oscillatory
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The series
a)
convergent
b)
divergent
c)
oscillatory
d)
None of these
|
|
Chirag Verma answered |
which is finite and non-zero.
Since, ∑vn is divergent, therefore, ∑un is also divergent.
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
a)
b)
c)
d)
|
|
Veda Institute answered |
Here, it is given that
Hence, ∑vn is divergent and ∑un is also divergent and
Hence, ∑vn is divergent and ∑un is also divergent and
The series 
convergent for- a)all real value o f x
- b)| x | < 1 only
- c)| x | ≤ 1
- d)-1 < x ≤ 1
Correct answer is option 'D'. Can you explain this answer?
The series convergent for
convergent fora)
all real value o f x
b)
| x | < 1 only
c)
| x | ≤ 1
d)
-1 < x ≤ 1
|
|
Veda Institute answered |
The series x 
is convergent.
If -1 <x ≤ 1.
is convergent.If -1 <x ≤ 1.
Find the sum of the series. 
- a)
(1/3)ln2−5/18 - b) (1/3)ln2−5/6
- c) (2/3)ln2−5/18
- d) (2/3)ln2−5/6
Correct answer is option 'C'. Can you explain this answer?
Find the sum of the series.
a)
(1/3)ln2−5/18
b)
(1/3)ln2−5/6
c)
(2/3)ln2−5/18
d)
(2/3)ln2−5/6
|
Pie Academy answered |
This problem is a very basic one, this problem can easily be solved by step by step solution. The steps are:
Step 1 : First we will ignore the summation part. We will factorize the denominator, because we are going step by step so our aim is to simplify the given problem first.
Step 2: After factorizing the the denominator we will reach to a position where we have to use partial fraction to go forward.
Step 3: In this step we will take care of the (−1)�� part, like how it will affect the series.
Step 4: After taking care of the (−1)�� we will now expand the summation (breaking it into infinite sum).
Step 5 : So after 4 steps we are halfway done now just the last simplification is left we will use the value
ln2=1−1/2+1/3−1/4+…
The series 
- a)divergent
- b)convergent
- c)unbounded
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The series
a)
divergent
b)
convergent
c)
unbounded
d)
None of these
|
|
Chirag Verma answered |
We have 
implies
Hence, series is convergent.
implies
Hence, series is convergent.
Let < an > and < bn> be two sequences of real numbers such that an = bn - bn + 1 for n = 1, 2 , 3 , 4 , . . . if Sbn is convergent, then which is always true?- a)San may not be convergent
- b)San is convergent and San = b1
- c)San is convergent and San = 0
- d)San is convergent and San = a1 - b1
- e)
Correct answer is option 'B'. Can you explain this answer?
Let < an > and < bn> be two sequences of real numbers such that an = bn - bn + 1 for n = 1, 2 , 3 , 4 , . . . if Sbn is convergent, then which is always true?
a)
San may not be convergent
b)
San is convergent and San = b1
c)
San is convergent and San = 0
d)
San is convergent and San = a1 - b1
e)
|
|
Chirag Verma answered |
The series 
- a)convergent
- b)divergent
- c)oscillatory
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The series
a)
convergent
b)
divergent
c)
oscillatory
d)
None of these
|
|
Chirag Verma answered |
The nth term of the given series is
Hence, ∑un is convergent.
Hence, ∑un is convergent.
The sequence 1, 1, 1, 1, 1…. is?- a)Absolutely summable
- b)Can’t say
- c)Is not absolutely summable
- d)None of the mentioned
Correct answer is option 'C'. Can you explain this answer?
The sequence 1, 1, 1, 1, 1…. is?
a)
Absolutely summable
b)
Can’t say
c)
Is not absolutely summable
d)
None of the mentioned
|
|
Chirag Verma answered |
For limit n tending to infinity the sum also tends to infinity and thus it is not summable.
The series 
- a)oscillating
- b)convergent
- c)unbounded
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The series
a)
oscillating
b)
convergent
c)
unbounded
d)
None of these
|
|
Chirag Verma answered |
or 0 as n is odd or even implies {Sn} - { 1 , 0 , 1 , 0 , . . . }It is an oscillating sequence
The series x 
is convergent, if
- a)0 <x < l/e
- b)x> l/e
- c)2/e <x< 3e
- d)3/e < x < 4/e
Correct answer is option 'A'. Can you explain this answer?
The series x is convergent, if
is convergent, ifa)
0 <x < l/e
b)
x> l/e
c)
2/e <x< 3e
d)
3/e < x < 4/e
|
|
Diksha Pandey answered |
Apply logarthmmic first test .
Let < Sn > and <tn> be two sequences such that 
and 
Then,- a)
- b)
does not exists finitely - c)
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
Let < Sn > and <tn> be two sequences such that and Then,
and Then,a)
b)
does not exists finitelyc)
d)
None of the above
|
|
Chirag Verma answered |
Since, we know that if
be a sequence such that 
and 
Hence,
and
be a sequence such that and Hence,
and
for all n ∈ N. Then the sequence {xn}The sequence {Sn} of real numbers given by Sn = 
is- a)a divergent sequence
- b)a Cauchy sequence
- c)an oscillatory sequence
- d)not a Cauchy sequence
Correct answer is option 'B'. Can you explain this answer?
The sequence {Sn} of real numbers given by Sn = is
a)
a divergent sequence
b)
a Cauchy sequence
c)
an oscillatory sequence
d)
not a Cauchy sequence
|
|
Veda Institute answered |
So <Sn> is monotonically increasing Next, we will show that it is bounded.
⇒ <Sn> is bounded. By the theorem's Every monotonic bounded sequence is convergent, Then <Sn> is convergent.
By the Lemma, if <Sn> is a convergent sequence of real numbers, Then <Sn> is a Cauchy-sequence.
The series 
convergent, if- a)|x |≤ 1
- b)|x|≤2
- c)|x| > 2
- d)| x | ≥ 4
Correct answer is option 'A'. Can you explain this answer?
The series convergent, if
convergent, ifa)
|x |≤ 1
b)
|x|≤2
c)
|x| > 2
d)
| x | ≥ 4
|
|
Chirag Verma answered |
Here given series is 
is convergent
∴ The given series will be convergent, if
Hence, vn is convergent series, so un is also convergent.
∴ The given series is convergent, if | x | ≤ 1.
is convergent∴ The given series will be convergent, if
Hence, vn is convergent series, so un is also convergent.
∴ The given series is convergent, if | x | ≤ 1.
is equal to- a)-π/4
- b)0
- c)π/4
- d)π/3
Correct answer is option 'C'. Can you explain this answer?
is equal toa)
-π/4
b)
0
c)
π/4
d)
π/3
|
|
Chirag Verma answered |
Here, it is given that
The interval of convergence of 
is- a)(–1, 1)
- b)[–1, 1)
- c)[-1, -1]
- d)[–1, 1]
Correct answer is option 'D'. Can you explain this answer?
The interval of convergence of is
a)
(–1, 1)
b)
[–1, 1)
c)
[-1, -1]
d)
[–1, 1]
|
|
Chirag Verma answered |
R = 1 / P = 1 = radius of convergence
When x = 1 & x = –1, Then
x = 1 & x = – 1.
so, its interval of convergence is exactly [–1, 1].
so, its interval of convergence is exactly [–1, 1].
The series whose nth term is 
- a)converges to the sum 0
- b)converges to the sum 1/2
- c)converges to the sum 1
- d)diverges
Correct answer is option 'D'. Can you explain this answer?
The series whose nth term is
a)
converges to the sum 0
b)
converges to the sum 1/2
c)
converges to the sum 1
d)
diverges
|
|
Chirag Verma answered |
Here it is given that
Now, taking auxiliary series
Which is finite and non-zero. Since, vn is divergent.
Hence, tn is also divergent.
Now, taking auxiliary series
Which is finite and non-zero. Since, vn is divergent.
Hence, tn is also divergent.
Which one of the following is incorrect?- a)every bounded sequence has a convergent subsequence
- b)every sequence has a monotonic subsequence
- c)every sequence has a limit point
- d)every sequence has a countable number of terms
Correct answer is option 'C'. Can you explain this answer?
Which one of the following is incorrect?
a)
every bounded sequence has a convergent subsequence
b)
every sequence has a monotonic subsequence
c)
every sequence has a limit point
d)
every sequence has a countable number of terms
|
|
Chirag Verma answered |
Solution :
The
set of all cluster points of a sequence is sometimes called the limit set. contains all but finitely many elements of the sequence). That is why we do not use the term limit point of a sequence as a synonym for accumulation point of the sequence.
converges, then 
is equal to
The radius of convergent of the series 
- a)∞
- b)zero
- c)1
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The radius of convergent of the series
a)
∞
b)
zero
c)
1
d)
None of these
|
|
Chirag Verma answered |
and the radius of convergence
The series 
is
- a)convergent
- b)divergent
- c)oscillatory
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The series is
isa)
convergent
b)
divergent
c)
oscillatory
d)
None of these
|
|
Chirag Verma answered |
Hence, by Cauchy’s root test, the series convergent.
In the given AP series the term at position 11 would be? 5, 8, 11, 14, 17, 20.........50. - a)45
- b)25
- c)15
- d)35
Correct answer is option 'D'. Can you explain this answer?
In the given AP series the term at position 11 would be? 5, 8, 11, 14, 17, 20.........50.
a)
45
b)
25
c)
15
d)
35
|
Ayush Gupta answered |
Explanation:
An arithmetic progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. In this case, the common difference is 3, as each term is obtained by adding 3 to the previous term.
Finding the 11th term:
To find the 11th term in the given AP series, we can use the formula for the nth term of an AP:
a_n = a + (n - 1)d
Where:
a_n is the nth term,
a is the first term,
n is the position of the term,
d is the common difference.
In this case:
a = 5 (the first term),
n = 11 (the position of the term),
d = 3 (the common difference).
Substituting the values into the formula:
a_n = 5 + (11 - 1) * 3
= 5 + (10) * 3
= 5 + 30
= 35
Therefore, the 11th term in the given AP series is 35.
Conclusion:
The 11th term in the given AP series is 35, so the correct answer is option D.
An arithmetic progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. In this case, the common difference is 3, as each term is obtained by adding 3 to the previous term.
Finding the 11th term:
To find the 11th term in the given AP series, we can use the formula for the nth term of an AP:
a_n = a + (n - 1)d
Where:
a_n is the nth term,
a is the first term,
n is the position of the term,
d is the common difference.
In this case:
a = 5 (the first term),
n = 11 (the position of the term),
d = 3 (the common difference).
Substituting the values into the formula:
a_n = 5 + (11 - 1) * 3
= 5 + (10) * 3
= 5 + 30
= 35
Therefore, the 11th term in the given AP series is 35.
Conclusion:
The 11th term in the given AP series is 35, so the correct answer is option D.
The third term of a geometric progression is 4. The product of the first five terms is
- a)43
- b)45
- c)44
- d) none of these
Correct answer is option 'B'. Can you explain this answer?
The third term of a geometric progression is 4. The product of the first five terms is
a)
43
b)
45
c)
44
d)
none of these
|
|
Chirag Verma answered |
here it is given that T3 = 4.
⇒ ar² = 4
Now product of first five terms = a.ar.ar².ar³.ar4
Now product of first five terms = a.ar.ar².ar³.ar4
= a5r10
= (ar2)5
= 45
= (ar2)5
= 45
The series 13 + 23 + 33 + .... is - a)divergent
- b)convergent
- c)bounded
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The series 13 + 23 + 33 + .... is
a)
divergent
b)
convergent
c)
bounded
d)
None of these
|
|
Chirag Verma answered |
{Sn} is increasing sequence and unbounded from above.
∴ It is divergent sequence.
Hence, the series is divergent series.
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