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Test: 35 Year JEE Previous Year Questions: Sequences and Series


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30 Questions MCQ Test Mathematics For JEE | Test: 35 Year JEE Previous Year Questions: Sequences and Series

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Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 1

If 1, log9 (31–x + 2), log3 (4.3x – 1) are in A.P. then x equals [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 1

(b) 1, log9 (31 – x + 2), log3 (4.3x – 1) are in A.P.
⇒ 2 log9  (31– x+2) = 1 + log3 (4.3x – 1) ⇒ log3 (31 – x + 2) = log33 + log3 (4.3x – 1)
⇒ log3 (31– x + 2) = log3 [3(4 × 3x – 1)] ⇒ 31– x + 2 = 3 (4.3x – 1) ⇒ 3.3–x + 2 = 12.3x – 3.
Put 3x = t

 = 

   Hence or  x = log33 – log3 4

⇒ x = 1 – log34

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 2

l, m, n are the pth, qth and rth term of a G. P. all positive, then   equals[2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 2

l = ARp –1 ⇒ log l = log A + (p – 1) log R
m = ARq –1 ⇒ log m = log A + (q – 1) log R
n = ARr –1 ⇒ log n = log A + (r – 1) log R
Now,

Operating  C1 - ( log R ) C2 + ( log R- logA) C3

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 3

The value of 21/4. 41/8. 81/16   ... ∞ is [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 3

The product is P = 21/4.22/8.23/16......... = 21/4 + 2/8 + 3/16+..........

Now let  ......(1)

......(2)

Subtracting (2) from (1)

or 

∴​ P = 2= 2

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 4

4. Fifth term of a GP is 2, then the product of its 9 terms is [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 4

ar4 = 2
a x ar x ar2x ar3 x ar4x ar5 x ar6 x ar7x ar8
= a9 r36 = (ar4)9 = 29 = 512

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 5

Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 5

Let a = first term of G.P. and  r = common ratio of G.P.; Then G.P. is a, ar, ar2

Given 

Also 

 a2 = 100(1 – r)(1 + r)... (ii)
From (i), a2 = 400(1 – r)2;
From (ii), we get 100(1 – r)(1 + r) = 400(1 – r)2
⇒ 1 + r = 4 – 4r
⇒ 5r = 3 ⇒ r = 3/5.

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 6

13 – 23 + 33 – 43 +...+93  = [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 6

13 – 23 + 33 – 43 + ........+ 93
= 13 + 23 + 33 +.......+ 93 – 2(23 + 43 + 63 + 83)

= 2025 – 1600 = 425

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 7

The sum of the series             [2003]

 up to ∞ is equal to

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 7

 

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 8

 is equal to

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 8

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 9

Let Tr be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers

  and  , then a – d equals [2004]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 9

............(1)

..............(2)

(1) - (2)  ⇒ (m-n)d = 

From (1)   a = 

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 10

The  sum of the first n terms of the series

 is       when n is even. When n is odd the sum is                      [2004]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 10

If n is odd, the required sum is

[∵ (n–1) is even
∴​ using given formula for the sum of (n–1)terms.]

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 11

If the coefficients of r th, (r + 1)th, and (r + 2)th terms in the the binomial expansion of (1 + y)m are in A.P., then m and r satisfy the equation [2005]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 11

Given mCr -1 , mCr , m Cr +1 are in A.P..

2m Cr = m Cr -1mCr+1

⇒ m2 - m(4r + 1) + 4r2 - 2=0 .

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 12

The sum of series         is [2004]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 12

We know that   

and 

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 13

IF  where a, b, c are in AA..PP and |a | < 1, | b | < 1, | c | < 1 then x, y, z ar e in [2005]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 13

a, b, c are in A.P.     OR 2b = a + c

 x, y, z are in H.P..

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 14

The sum of the series [2005]

.................ad inf. is

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 14

Putting x =  we get

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 15

Let a1,a2 ,a3 ............ be terms on A.P. If then  equals [2006]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 15

For 

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 16

 If a1 , a2 , ........,an are in H.P., then the expression a1a2 + a2 a3 + ..........+ an -1an is equal to [2006]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 16

Also,

Which is the required result.

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 17

The sum of series ... upto infinity is [2007]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 17

We know that ex =  

Put x = – 1

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 18

In a geometric progression consisting of positive terms, each term equals the sum of the next two  terms. Then the common ratio of its progression is equals [2007]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 18

Let the series a, ar, ar2, ..... are in geometric progression. given, a = ar + ar2

[∵ terms of G.P. are positive
∴ r should be positive]

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 19

The first two terms of a geometric progression add up to 12. the sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is [2008]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 19

As per question,

a + ar = 12 …(1)

ar2 + ar3 = 48 …(2)

(Qterms are = + ve and –ve alternately)

⇒ a = –12

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 20

The sum to infinite term of the series       [2009]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 20

We have

        ....(1)

Multiplying both sides by  we get

....(2)

Subtracting eqn. (2) from eqn. (1) we get

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 21

A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth minute. If a1 = a2 = ... = a10 = 150 and a10, a11, ... are in an AP with common difference –2, then the time taken by him to count all notes is [2010]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 21

Till 10th minute number of counted notes = 1500

n2 - 149n + 3000= 0

⇒ n = 125, 24

But n = 125 is not possible

∴ total time = 24 + 10 = 34 minutes.

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 22

A man saves 200 in each of the first three months of his service. In each of the subsequent months his saving increases by 40 more than the saving of immediately previous month. His total saving from the start of service will be 11040 after [2011]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 22

Let required number of months = n

∴ 200 × 3 + (240 + 280 + 320 + ... +  (n – 3)th term) = 11040

⇒ (n - 3)[240 + 20n - 80]= 10440

⇒ (n - 3)(20n + 160)= 10440

⇒ (n - 3)(n + 8)= 522

⇒ n2 + 5n - 546=0

⇒  (n + 26) (n – 21) = 0

∴ n = 21

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 23

Statement-1: The sum of the series 1 + (1 + 2 + 4) +  (4 + 6 + 9) + (9 + 12 + 16) + .... + (361 + 380 + 400) is 8000

Statement-2:    for any natural number n. [2012] 

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 23

nth term of the given series

= Tn = (n - 1)2 +( n -1) n+n2

⇒ n = 20 which is a natural number.
Now, put n =1,2,3,....20

Hence, both the given statements are true and statement 2 supports statement 1.

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 24

The sum of first 20 terms of the sequence 0.7, 0.77, 0.777,....., is [JEE M 2013]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 24

Given sequence can be written as

Multiply and divide by 9

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 25

If (10)9 + 2(11)1 (108) + 3(11)2 (10)+ K........  +10 (11)= k (10)9, then k is equal to: [JEE M 2014]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 25

Let 109 + 2. (11)(10)8 + 3(11)2 (10)7 + ... + 10(11)9 = k(10)9

Let x = 109 + 2.(11)(10)8 + 3(11)2(10)7 + ... + 10(11)9

Multiplied by  on both the sides

⇒ x = 1011 = k.109 Given   ⇒ k = 100

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 26

Three positive numbers form an increasing G. P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is: [JEE M 2014]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 26

Let a, ar, ar2 are in G.P.
According to the question a, 2ar, ar2 are in A.P.
⇒ 2 × 2ar = a + ar2
⇒ 4r = 1 + r2 ⇒ r2 – 4r + 1 = 0

Since r > 1

is rejected

Hence, 

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 27

The sum of first 9 terms of the series.     [JEE M 2015]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 27

nth term of series =

Sum of n term  = 

Sum of 9 terms

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 28

If m is the A.M. of two distinct real numbers l and n(l, n > 1) and G1, G2 and G3 are three geometric means between l and n, then   equals : [JEE M 2015]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 28

 and common ratio of G.P. = r =


= ln (l + n)2
= ln × 2m2
= 4lm2n

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 29

If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is : [JEE M 2016]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 29

Let the GP be a, ar and ar2 then a = A + d; ar = A + 4d; ar2 = A + 8d

Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 30

If th e sum of th e fir st ten ter ms of th e series

then m is equal to : [JEE M 2016] 

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Sequences and Series - Question 30

⇒ m= 101.

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