JEE  >  Subjective Type Questions: Sequences and Series | JEE Advanced

# Subjective Type Questions: Sequences and Series | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

## Document Description: Subjective Type Questions: Sequences and Series | JEE Advanced for JEE 2022 is part of Maths 35 Years JEE Main & Advanced Past year Papers preparation. The notes and questions for Subjective Type Questions: Sequences and Series | JEE Advanced have been prepared according to the JEE exam syllabus. Information about Subjective Type Questions: Sequences and Series | JEE Advanced covers topics like and Subjective Type Questions: Sequences and Series | JEE Advanced Example, for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Subjective Type Questions: Sequences and Series | JEE Advanced.

Introduction of Subjective Type Questions: Sequences and Series | JEE Advanced in English is available as part of our Maths 35 Years JEE Main & Advanced Past year Papers for JEE & Subjective Type Questions: Sequences and Series | JEE Advanced in Hindi for Maths 35 Years JEE Main & Advanced Past year Papers course. Download more important topics related with notes, lectures and mock test series for JEE Exam by signing up for free. JEE: Subjective Type Questions: Sequences and Series | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
 1 Crore+ students have signed up on EduRev. Have you?

Q.1. The harmonic mean of two numbers is 4. Their arithmetic mean A and the geometric mean G satisfy the relation. 2A + G= 27
Find the two numbers. (1979)

Ans. 3 and 6 or 6 and 3

Sol. Let the two numbers be a and b, then

Also 2A + G2 = 27 ⇒ a + b + ab = 27 ....(2)

Putting ab = 27 – (a+b) in eqn. (1), we get

a + b = 9 then ab = 27 –  9 = 18

Solving the two we get a = 6, b = 3 or a = 3, b = 6, which are the required numbers.

Q.2. Th e in ter ior an gles of a polygon ar e in ar ith metic progression. The smallest angle is 120°, and the common difference is 5°, Find the number of sides of the polygon. (1980)

Ans. 9

Sol. Let there be n sides in the polygon.
Then by geometry, sum of all n interior angles of polygon = (n – 2) × 180° Also the angles are in A.P. with the smallest angle = 120° , common difference = 5° ∴ Sum of all interior angles of polygon

Thus we should have

⇒ 5n2 + 235n= 360n- 720
⇒ 5n2 - 125n + 720= 0
⇒ n- 25n + 144= 0
⇒ (n -16) (n - 9)= 0
⇒ n = 16,  9

Also if n = 16 then 16th angle = 120 + 15 × 5 = 195° > 180°

∴ not possible. Hence n = 9.

Q.3. Does there exist a geometric progression containing 27, 8 and 12 as three of its terms  ? If it exits, how many such progressions are possible ? (1982 - 3 Marks)

Ans. yes, infinite

Sol. If possible let for a G.P.
Tp= 27 = ARp–1 ....(1)
Tq= 8 = ARq–1 ....(2)
Tr= 12 = ARr–1 ....(3)
From (1) and (2)

....(4)

From (2) and  (3):

....(5)

From (4) and (5):

R = 3/2; p – q = 3 ;   q – r = – 1 p – 2q + r = 4;     p, q, r ∈ N ....(6)

As there can be infinite natural numbers for p, q and r to satisfy equation (6)

∴ There can be infinite G.P’s.

Q.4. Find three numbers a, b, c, between 2 and 18 such that (i) their sum is 25 (ii) the numbers 2, a, b sare consecutive terms of an A.P. and (iii) the numbers b, c, 18 are consecutive terms of a G.P. (1983 - 2 Marks)

Ans.  a = 5,  b = 8,  c = 12

Sol. 2 < a, b, c  <  18         a + b + c = 25 ....(1)

2, a, b are in AP ⇒ 2a = b + 2 ⇒ 2a – b = 2 ....(2)

b, c, 18 are in GP ⇒ c2 = 18b ....(3)

From  (2)

+

(3) ⇒ c2 = 6 (48 - 2c) ⇒ c+ 12c - 288= 0

⇒ c = 12, – 24 (rejected)   ⇒ a = 5,  b = 8,  c = 12

Q. 5. If a > 0, b > 0 and c > 0, prove that

(1984 - 2 Marks)

Ans. Sol. Given that a, b, c > 0

We know for +ve numbers A.M. ≥ G.M.

∴ For +ve numbers a, b, c we get

....(1)

Also for +ve numbers  we get

....(2)

Multiplying in eqs (1) and (2) we get

Proved.

Q.6. If n is a natural number such that

an d p1, p2, ....., pk are distinct primes, then show that ln n ≥ k ln2 (1984 - 2 Marks)

Ans. Sol. Given that          ....(1)

Where n∈N and  p1,  p2, p3, ......pare distinct prime numbers.
Taking log on both sides of eq. (1),

we get log n = α1 log p1+ α2 log p2 + ....+ ak  log pk ....(2)

Since every prime number is such that

p≥ 2

....(3)

∀ i= 1 (1) k Using (2) and (3)

we get log n ≥ α1 log 2 + αlog 2 + α3 log 2 + ....+ αk log 2
⇒ log n ≥ (α12+ ....+ αk) log 2
⇒ log n ≥ k log 2 Proved.

Q.7. Find the sum of the series :

... up to  terms]

Ans. Sol. The given series is

Now,

Similarly,

Hence the given series is,

.... to terms

[ Summing the G.P.]

Q.8. Solve for x the following equation : (1987 - 3 Marks)
log(2 x +3) (6 x2 + 23x + 21) = 4 - log (3x+7) (4 x2 + 12x+ 9)

Ans.  -1/4

Sol. The given equation is log (2x + 3) (6x2 + 23 + 21)
= 4 – log3x+7 (4x2 + 12x + 9)

⇒ log(2x+3) (6x2 + 23x + 21) + log(3x+7) (4x2 + 12x + 9) = 4

⇒ log(2x+3) (2x + 3) (3x + 7) + log(3x+7) (2x + 3)2  x =4

⇒ 1+log(2x+3) (3x + 7) + 2log(3x+7) (2x + 3) =4

⇒ log(2x+3)(3x + 7) +

Let log(2x+3)(3x + 7) = y ....(1)

y2 -3y+2=0

⇒ (y – 1)  (y – 2) = 0 ⇒ y = 1, 2

Substituting the values of y in (1), we get

⇒ log(2x+3)(3x + 7) = 1  and     log(2x+3)(3x + 7) = 2

⇒ 3x + 7 = 2x + 3 an d3x + 7 = (2x + 3)

⇒ x  = – 4 an d4x2 + 9x + 2 = 0

⇒ x = – 4 an d(x + 2) (4x + 1) = 0

⇒ x = – 4 and x = – 2, x = –

As log a x is defined for x > 0 and a > 0 (a ≠ 1), the possible value of x should satisfy all of the following inequalities :

⇒ 2x + 3 > 0 and 3x + 7 > 0

Also 2x + 3 ≠ 1 and 3x + 7 ≠ 1

Out of x  =  – 4, x = – 2 an d x  = – only x  = –

satisfies the above inequalities.

So only solution is x = –

Q. 9. If log 3 2 , log3 (2 x , 5) , and are in arithmetic progression, determine the value of x. (1990 -  4 Marks)

Ans.  3

Sol. Given that log2, log3 (2– 5), log3 (2x –7/2) are in A.P.
⇒ 2 log3 (2x – 5) = log3 + log3 (2x – 7/2)

⇒ (2 x - 5) 2

⇒ (2x)2 – 10.2+ 25 – 2.2x + 7 = 0 ⇒ (2x)2 – 12.2+ 32 = 0

Let 2x  = y, then we get, y2 – 12y + 32 = 0
⇒ (y – 4) (y – 8) = 0
⇒ y = 4  or  8
⇒ 2x = 22  or 23 ⇒ x = 2  or 3

But for log(2x – 5) and log3 (2– 7/2) to be defined
2x –  5 > 0 and 2x  –  7/2 > 0 ⇒ 2x  >  5 and 2 > 7/2
⇒ 2> 5 ⇒ x ≠ 2  and therefore x = 3.

Q.10. Let p be the first of the n arithmetic means between two numbers and q the first of n harmonic means between the same numbers. Show that q does not lie between p and

(1991 -  4 Marks)

Ans. Sol. Let a and b be two numbers and A1, A2, A3, .... An be n A.M’s between a and b.
Then a, A1, A2, ..... An, b are in A.P.
There are (n + 2) terms in the series, so that

a + (n + 1) d = b

....(1)

The first H.M. between a and b, when nHM’s are inserted between a and b can be obtained by  replacing a by   and b by   in eq. (1) and then taking its reciptocal.

Therefore,

...(2)

We have to prove that q cannot lie between p

Now, n + 1>n –1

....(3)
Now to prove the given, we have to show that q is less than p.

For this, let,

⇒ (provided a and b and hence p and q are +ve) p > q            ....(4)

From 3 and (4), we get,

∴ q can not lie between p and  p, if a and b are +ve numbers.

Q.11. If S1, S2 ,S3, ............... , Sn are the sums of infinite geometric series whose first terms are 1, 2, 3, ..............., n and whose common ratios are

respectively,,then find the values of  (1991 -  4 Marks)

Ans. Sol.  We have,

....................................................
....................................................

NOTE THIS STEP :

Q.12. The real numbers x1, x2, x 3 s atis fying the equation x, x2 + βx + γ < 0 are in AP. Find the intervals in which β and γ lie. (1996 - 3 Marks)

Ans. Sol. Since x1,  x2,  x3 are in A.P.
Therefore, let  x= a – d, x= a and x3 = a + d and x1,  x2,  x3 are the roots of  x3 – x2 + βx  + γ =0
We have ∑α = a – d + a + a + d = 1 ....(1)
∑αβ = (a – d) a + a ( a + d) + (a – d) (a + d) = b ....(2)
αβγ = (a – d) a (a + d) = – γ ....(3)

From (1), we get, 3a = 1 ⇒ a = 1/3

From (2), we get, 3a2 – d2 = β

⇒ 3(1/3)2 – d2 = β ⇒ 1/3 – β = d

We know that d> 0 ∀ d ∈ R

∵ d 2 ≥ 0

From (3), a (a2 – d2)  = – γ

Q.13. Let a, b, c, d be real numbers in G.P. If u, v, w, satisfy the system of equations (1999 - 10 Marks)

then show that the roots of the equation

and 20x+ 10 (a - d)2 x - 9 = 0 are reciprocals of each other.

Ans.

Sol. Solving the system of equations, u + 2v + 3w = 6,

4u + 5v + 6w = 12 and 6u + 9v = 4

we get u = – 1/3, v = 2/3, w = 5/3

Let r be the common ratio of the G.P., a, b, c, d. Then b = ar, c = ar2,  d = ar3 .
Then the first equation

+ [(b – c)+ (c – a)+ (d – b)2]x + (u + v + w) = 0 becomes

The second equation is, 20 x2 + 10 (a – ar3)2  x – 9 = 0

i.e., 20 x2 + 10 a (1 – r3)x – 9 = 0 ....(2)

Since (2) can be obtained by the substitution x → 1/x , equations (1) and (2) have reciprocal roots.

Q.14. The fourth power of the common difference of an arithmatic progression with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.           (2000 - 4 Marks)

Ans.

Sol.  Let a – 3d, a – d, a + d and a + 3d  be any  four consecutive terms of an A.P. with common difference 2d.

∵ Terms of A.P. are integers, 2d is also an integer.
Hence  P = (2d)4 + (a – 3d) (a – d) (a + d) (a + 3d)

=16 d 4 + (a 2 – 9 d 2) (a 2 – d 2) = (a 2 – 5 d 2)2

Now, a 2 – 5 d 2 = a 2 – 9 d 2 + 4 d 2 = (a – 3 d) (a + 3 d) + (2 d)2 =  some integer

Thus, P = square of an integer.

Q.15. Let a1, a2, …, an be positive real numbers in geometric progression. For each n, let An, Gn, Hbe respectively, the arithmetic mean, geometric mean, and harmonic mean of a1, a2, …, an. Find an expression for the geometric mean of G1, G2, …, Gn in terms of A1, A2 , …, An, H1, H2, …, Hn. (2001 - 5 Marks)

Ans.

Sol. Given that a1, a2, ....an are +ve real no’s in G.P.

An is A.M. of a1, a2, ....., an

.... (1)  (For r ≠ 1)

Gn is G.M. of a1, a2, ...., an

.... (2)   (r ≠ 1)

His H.M. of a1,  a2, ....., an

(r ≠ 1) ...(3)
We also observe that

∴ AHn =Gn2....(4)

∴ Now, G.M. of   G1, G2, .... Gn is

[Using equation (4)

G = (A1A2....AnH1H2....Hn)1/2n   ...(5)

If  r = 1 then An = Gn = Hn = a

Also An H= Gn2

∴ For r = 1 also, equation (5) holds.

Hence we get, G = (A1A2....AnH1H2....Hn)1/2n

Q.16. Let a, b be positive real numbers. If a, A1, A2, b are in arithmetic progression, a, G1, G2, b are in geometric progression and a, H1, H2, b are in harmonic progression,

show that      (2002 - 5 Marks)

Ans.

Sol. Clearly A1 + A= a + b

Q.17. If a, b, c are in A.P., a2, b2, c2 are in H.P., then prove that either a = b = c or a, b,  form a G.P.. (2003 - 4 Marks)

Ans.

Sol. Given that a, b, c are in A.P.
⇒ 2b = a + c ....(1)
and a2, b2, care in H.P.

⇒ ac2 + bc= a2b + a2c     [∵ a – b = b – c]
⇒ ac (c – a) + b (c – a) (c + a) = 0
⇒ (c – a) (ab + bc + ca) = 0
⇒ either c – a = 0 or ab + bc + ca = 0

⇒ either c = a or (a + c) b + ca = 0 and then from (i)
2b2 + ca = 0

Either a = b = c

i.e. a, b, – c/2 are in G.P. Hence Proved.

Q.18. If anand

bn = 1 – an, then find the least natural number n0 such that

(2006 - 6M)

Ans.

Sol.

⇒ (– 3)n +1 < 22n–1 For n to be even, inequality always holds. For n to be odd, it holds for n ≥ 7.
∴ The least natural no., for which it holds is 6
(∵ it holds for every even natural no.)

The document Subjective Type Questions: Sequences and Series | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE is a part of the JEE Course Maths 35 Years JEE Main & Advanced Past year Papers.
All you need of JEE at this link: JEE

## Maths 35 Years JEE Main & Advanced Past year Papers

132 docs|70 tests
 Use Code STAYHOME200 and get INR 200 additional OFF

## Maths 35 Years JEE Main & Advanced Past year Papers

132 docs|70 tests

### Top Courses for JEE

Track your progress, build streaks, highlight & save important lessons and more!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;