ch - 9 sequences and series - badhan education 11th/class 11th... · 5 corresponding series = 3 +...

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1. Write the first five terms of each of the sequences. whose nth terms are an = n(n+2)

• Solution: an = n(n+2)

For n = 1, a1 = 1(1 + 2) = 3

For n = 2, a2 = 2(2 + 2) = 8

For n = 3, a3 = 3(3 + 2) = 15

For n = 4, a4 = 4(4 + 2) = 24

For n = 5, a5 = 5(5 + 2) = 35

Thus first five terms are 3, 8, 15, 24, 35.

2. Write the first five terms of each of the sequences. whose nth terms

are an =

• Solution:

an =

For n = 1, a1 = =

For n = 2, a2 = =

For n = 3, a3 = =

For n = 4, a4 = =

For n = 5, a5 = = Hence first 5 terms of the given sequence are

3. Write the first five terms of each of the sequences. whose nth terms are an = 2n

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• Solution: an = 2

n

For n = 1, a1 = 21 = 2

For n = 2, a2 = 22 = 4

For n = 3, a3 = 23 = 8

For n = 4, a4 = 24 = 16

For n = 5, a5 = 25 = 32

Hence first five terms of the given sequence are 2, 4, 8, 16 and 32.

4. Write the first five terms of each of the sequences. whose nth terms are

an =

• Solution:

an = Substituting n = 1, 2, 3, 4 and 5 we get

a1 = (2 x 1 - 3)/6 = -1/6 a2 = (2 x 2 - 3)/6 = (4 - 3)/6 = 1/6

a3 = (2 x 3 - 3)/6 = (6 - 3)/6 = 3/6 = ½

a4 = (2 x 4 - 3)/6 = (8 - 3)/6 = 5/6

a5 = (2 x 5 - 3)/6 = (10 - 3)/6 = 7/6 Therefore the first term is –1/6, 1/6, ½, 5/6 and 7/6.

5. Write the first five terms of each of the sequences. whose nth terms are an = (-

1)n-1 5n+1

• Solution: an = (-1)

n-1 5n+1

Substituting n = 1, 2, 3, 4 and 5 we get

a1 = (-1)1-1 51+1= (-1)0 52 = 25

a2 = (-1)2-1 52+1= (-1)1 53 = -125

a3 = (-1)3-1 53+1= (-1)2 54 = 625







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a4 = (-1)4-1 54+1= (-1)3 55 = -3125

a5= (-1)5-1 55+1= (-1)4 56 = 15625

Therefore the first term is 25, -125, 625, -3125 and 15625.

6. Write the first five terms of each of the sequences. whose nth terms are an =

n(n2 + 5)/4

• Solution: an = n(n

2 + 5)/4 Substituting n = 1, 2, 3, 4 and 5 we get

a1 = 1(12 + 5)/4 = 6/4 = 3/2

a2 = 2(22 + 5)/4 = 2(4 + 5)/4 = 18/4 = 9/2

a3 = 3(3

2 + 5)/4 = 3(9 + 5)/4 = 135/2 a4 = 4(4

2 + 5)/4 = 4(16 + 5)/4 = 84/4 = 21 a5 = 5(5

2 + 5)/4 = 5(25 + 5)/4 = 150/2 = 75 Therefore the first term is 3/2, 9/2, 135/2, 21 and 75/2.

7. Find the indicated terms in each of the sequences. whose nth terms are an = 4n

– 3, a17, a24

• Solution: For n = 17, a17 = 4 × 17 – 3 = 68 – 3 = 65

For n = 24, a24 = 4 × 24 – 3 = 96 – 3 = 93

Hence a17 = 65 and a24 = 93

8. Find the indicated terms in each of the sequences. whose nth terms are

an =

• Solution:

an =







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for n = 7, a7 = =

Hence a7 =

9. Find the indicated terms in each of the sequences. whose nth terms are an = (-

1)n – 1n3, a9

• Solution: an = (-1)

n – 1. n3 for n = 9, a9 = (-1)

9 – 1. (9)3 = 9 × 9 × 9 = 729 Hence an = 729

10. Find the indicated terms in each of the sequences. whose nth terms are

an =

• Solution:

an =

For N = 20, a20 = 20 = =

Hence a20 =

11. Write the first five terms of each of the sequence in obtain the corresponding series

a1 = 3 an = 3an – 1 + 2, for all n > 1

• Solution: a1 = 3

an = 3an – 1 + 2, for all n > 1

For n = 2, a2 = 3an – 1 + 2, for all n > 1

= 3a1 + 2 = 3 × 3 + 2 = 11 For n = 3, a3 = 3a2 + 2 = 3 × 11 + 2 = 35 For n = 4, a4 = 3a3 + 2 = 3 × 35 + 2 = 107 For n = 5, a5 = 3a4 + 2 = 3 × 107 + 2 = 323 Hence the required sequence is

3, 11, 35, 107, 323+……..







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Corresponding series

= 3 + 11 + 35 + 107 + 323+……..

12. Write the first five terms of each of the sequence in obtain the corresponding

series a1 = -1, an = an-1/n, (n ≥≥≥≥ 2)

• Solution: a1 = -1

an =

For n = 2, a2 =

For n = 3, a3 = = -

For n = 4, a4 = = -

For n = 5, a5 = = - Hence first 5 terms are

-1, - and the series is

(-1) +

= -1 -

13. Write the first five terms of each of the sequence in obtain the corresponding

series a1 = a2 = 2, an = an-1 – 1, (n > 2)

• Solution: a1 = a2 = 2 and an = an – 1, n > 2 For n = 3, a3 = a2 – 1 = 2 – 1 = 1 For n = 4, a4 = a3 – 1 = 1 – 1 = 0 For n = 5, a5 = a4 – 1 = 0 – 1 = -1 Hence first 5 terms are 2, 2, 1, 0, -1 and the corresponding series = 2 + 2 + 1 + 0 +(-1)+…

14. The Fibonacci sequence is defined by a1 = 1 = a2 , an = an-1 + an-2

(n > 2). Find an+1/an, for n = 1, 2, 3, 4, 5.

• Solution: Substituting n = 3, 4, 5 and 6 in an = an-1 + an-2







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a3 = a3-1 + a3-2 = a2 + a1 = 1 + 1 = 2 a4 = a4-1 + a4-2 = a3 + a2 = 2 + 1 = 3 a5 = a5-1 + a5-2 = a4 + a3 = 3 + 2 = 5 a6 = a6-1 + a6-2 = a5 + a4 = 5 + 3 = 8 Substituting n = 1

Substituting n = 2, 3, 4 and 5 we get

The required series is 1, 2, 3/2, 5/3 and 8/5.

15. Find the sum of odd integers from 1 to 2001.

• Solution: Sn = 1 + 3 + 5 + 7 + ….. + 2001

Now a = 1

d = 3 – 1 = 2, l = 2001

Let tn = 2001

Therefore a + (n + 1) d = 2001

or 1 + (n – 1)(2) = 2001

or (2n – 1) = 2000

or n – 1 = 1000

or n = 1001

We have Sn =







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or S1001 =

= = 1001 × 1001 = 1002001

16. Find the sum of all natural numbers between 100 and 1000 which are

multiples of 5.

• Solution: The next number to 100 which is multiple of 5 is 105. Similarly the next number to 105 multiple of 5 is 110. The greatest number multiple of 5 but less than 1000 is 995.

a = 105, d = 110 – 105 = 5

an = a + (n - 1)d

995 = 105 + (n - 1)5

995 = 105 + 5n – 5

5n = 895

n = 179

Son =

=

= 98450 Therefore the sum of all natural numbers between 100 and 1000 are 98450.

17. In an A.P., the first term is 2 and the sum of the first five terms is one-fourth

of the sum of the next five terms. Show that the 20th term is –112.

• Solution: a = 2, S5 = (1/4) (S10 - S5)

S5= (5/2)(2×€2 + 4d) = (5/2)(4 + 4d) = 10 + 10d S10 = (10/2)(2×€2 + 9d) = (10/2)(4 + 9d) = 5(4 + 9d) = 20 + 45d Since S5 = (1/4) (S10 - S5)







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10 + 10d = (1/4)(20 + 45d – 10 - 10d)

40 + 40d = 10 + 35d

-5d = 30 d = -6

a20 = 2 + (20 –1)(-6)

= 2 + (20 –1)(-6)

= 2 – 114

= - 112

18. How many terms of the A.P. –6, -11/2, -5, ….. are needed to give the sum –

25?

• Solution: a= -6, d = (-11/2) + 6 = (-11 + 12)/2 = ½, Sn = -25 -25 = (n/2)[2(-6) + (n -1)(1/2)]

-25× 4 = n(-24 + n –1)

- 100 = n(-25 + n)

n2 -25n + 100 = 0

n2 -20n – 5n + 100 = 0

n(n – 20) – 5(n – 20) = 0

(n – 5)(n – 20) = 0

n = 5, n = 20 Therefore the required number of terms is 5 or 20.

19. If the pth term of an A.P is and the qth term is , prove that the sum of the

first pq terms must be (pq + 1).

• Solution:

tp = a + (p-1)d = ……….(1)







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tq = a + (q-1)d = ……….(2)

(1) - (2)

a + (p-1)d =

a + (q-1)d = ---------------------

(p - q)d = ----------------------

d = =

a + (p-1) =

a = - = =

Sum of the 1st pq terms = ]

= =

Spq =

20. If the sum of a certain number of terms of the A.P. 25, 22, 19,….. is 116, find

the last term.

• Solution: Given. A.P.: 25, 22, 19,…..

Here Solution = 116, a = 25, d = 22 – 25 = -3

Let n be the number of terms.

i.e., Last term = tn

Now Sn =







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or 116 =

or 232 = n[50 – 3n + 3]

or n[53 – 3n] = 232

or 3n2 – 53n + 232 = 0

or 3n2 – 29n – 24n + 232 = 0

or n(3n – 29) –8(3n – 29) =0 or (3n – 29) (n – 8) =0

or n = 8 as n ∑ N

Now Last term = Tn = a+(n – 1) d

or T8 = 25 + (8 – 1)(-3)

= 25 – 21 = 4 Therefore the last term is 4.

21. Find the sum to n terms of the A.P. whose kth term is 5k + 1.

• Solution: Tk = 5k + 1

T1 = 5 × 1 + 1 = 6

T2 = 5 × 2 + 1 = 11

T3 = 5 × 3 + 1 = 16

T4 = 5 × 4 + 1 = 21

Now a = 6, d = 11 – 6 = 5

Thus Sn =

=

=







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= =

22. If the sum of n terms of an A.P is (pn + qn2), where p and q are constants,

find the common difference.

• Solution: For the A.P.

Sn = pn + qn2

Now S1 = p × 1 + q(1)2

S1 = p × q ⇒ T1 = p + q

Also S2 = p × 2 + q(2)2

= 2p + 4q We have T1 + T2 = 2p + 4q

or T2 = 2p + 4q – T1

or T2 = 2p + 4q – (p + q) or T2 = p + 3q Hence common difference = T2 – T1

= p + 3q –(p + q)

= p + 3q – p – q = 2q

23. If the sum of n terms of two arithmetic progressions are in the ratio 5n + 4 :

9n + 6, find the ratio of their 18th terms.

• Solution: For first A.P. : Let first term = a1

Common difference = d1

and Sn1 = Sum of n terms

For second A.P. :

Let first term = a2

Common difference = d2







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and Sn2 = Sum of n terms

Given =

Required to find

=

We have =

or =

or =

Taking = 17 or n = 34 + 1 = 35, we have = =

24. If the sum of first p terms of an A.P. is equal to the sum of the first q terms,

then find the sum of the first(p+q) terms.

• Solution: Given. For the A.P.

Sp = Sq

Let the first term = a, common difference = d

Therefore, according to the question = Required. To find Sp + q

Now =







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According to question

or p[2a + pd – d] = q[2a + qd – d]

or 2ap + p2d – pd = 2aq + q2d – qd or 2a(p - q)2 +d(p2 – q2) – d(p –q) = 0

or 2a + d(p + q) - d = 0

2a + (p + q – 1)d = 0

Now Sp + q =

= = 0

25. Sums of the first p, q, r terms of an A.P are a, b, c respectively. Prove

that

• Solution: Let x be the first term and d the common difference.

Sp = p/2[ 2x + (p-1)d] = a

Sq = q/2[2x + (q-1)d] = b

Sr = r/2[2x + (r-1)d] = c

2x + (p-1)d = 2a/p………..(i)

2x + (q-1)d = 2b/q………..(ii)

2x + (r-1)d = 2c/r………….(iii)

Eliminating x and d from (i), (ii) and (iii).

(p-q)d = 2a/p - 2b/q [Subtracting (i) and (ii)] ……(iv)

(q-r)d = 2a/q - 2c/r [Subtracting (iii) and (ii)] … (v)

[ Dividing (iv) and (v)]







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=

26. The ratio of the sums of m and n terms of an A.P is m2:n2. Show that the ratio

of mth and nth term is 2m – 1: 2n – 1.

• Solution: Sm : Sn = m

2:n2

Sm : m2 = Sn : n

2 = k

Tm : Tn = Sm – Sm - 1 : Sn – Sn – 1

= k{m2 – (m - 1)2} : k{n2 – (n - 1)2} = 2m – 1 : 2n - 1

27. If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the

value of m.

• Solution: Given. Sn = 3n

2 + 5n

S1 = 3(1)2 + 5 × 1 = 8 ⇒ T1 = 8

S2 = 3(2)2 + 5 × 2 = 22

Thus T1 + T2 = 22

or 8 + T2 = 22

or T2 = 14







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Now a = 8, d = 14 – 8 = 6

As Tm = 164 or a +(m – 1)d = 164 or 8 + (m – 1) (6) = 164 or 8 + 6m – 6 = 164 or 6m = 162

or m = = 27 Hence m = 27

28. Insert five number between 8 and 26 such that the resulting sequence is an

A.P.

• Solution: Let the five A.M’s be A1, A2, A3, A4, A5

Therefore 8, A1, A2, A3, A4, A5, 26 are in A.P. Let d be the common difference

Here T1 = 8

T7 = 26

⇒€a + ( 7 – 1)d = 26

or a + 6d = 26

or 8 + 6d = 26

or 6d = 18

or d = 3

Therefore A1 = 8 + 3 = 11 A2 = 11 + 3 = 14 A3 = 14 + 3 = 17 A4 = 17 + 3 = 20 A5 = 20 + 3 = 23

Thus five required numbers are

11, 14, 17, 20 and 23

29. If (an + bn)/(an-1 + bn-1) is the A.M. between a and b, then find the value of n.

• Solution: The A.M between a and b is (a + b)/2. (an + bn)/(an-1 + bn-1) = (a + b)/2







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2(an + bn) = (a + b)(an-1 + bn-1)

2an + 2bn = an + bn + ban-1 + abn-1

an + bn - ban-1 - abn-1 = 0

an - ban-1 + bn - abn-1 = 0

an-1(a – b) - bn-1(b – a) = 0

(an-1 - bn-1)(b – a) = 0

(an-1 - bn-1) = 0 or (b – a) = 0

an-1 = bn-1 or b = a

If a = b, then n can take any value.

If a = b, then an-1 = bn-1 = an-1/bn-1 = 1 = (a/b)n-1 = 1

As a = b, a/b = 1.

Therefore, (a/b)n-1 = 1 = n – 1 = 0 = n = 1

30. Between 1 and 31, m arithmetic means have been inserted in such a way that

the ratio of the 7th and (m - 1)th means is 5:9. Find the value of m.

• Solution: Let the means be x1, x2 , x3 , x4 ,……, xm.

So that 1, x1, x2, x3, x4,……, xm, 31 is an A.P of m + 2 terms am + 2 = a + (m + 1)d 31 = 1 + (m + 1)d (m + 1)d = 30 d = 30/(m + 1) ………………….(i) x7 : xm - 1 = 5:9

a + 7d : a + (m - 1)d = 5 : 9 9(1 + 7d) = 5(1 + (m - 1)d)

9 + 63d = 5 + 5md - 5d

4 = -63d + 5md - 5d 4 = (-68 + 5m)d







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4 = (-68 + 5m)[30/ (m +1)]

4(m + 1) = (-68 + 5m)30

2m + 2= -1020 + 75m 73m = 1022

m = 14

31. A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs.5 every month, what amount will he pay in the 30th instalment.

• Solution: Thus he pays Rs.100, Rs.105, Rs.100……

Now a = 100, d = 5

To find, T30

We know that

Tn = a + (n - 1)d T30 = 100 + (30 - 1) × 5 = 100 + 145 = 245

32. The difference between any two consecutive interior angles of a polygon is 50.

If the smallest angle is 1200, find the number of the sides of the polygon.

• Solution: We know that if the number of sides is n, then the sum of interior angles of a polygon with n sides = (2n - 4) right angles = (180n- 360) degrees Now the sequence of angles

120°, 125°, 130°,... form an A.P. with a = 120°, d = 5°

Sn =

⇒ 180n – 360 = or 2(180n – 360) = n[240 + 5n - 5)]







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or 360n - 720 = 240n + 5n2 + 5n

or 5n2 – 125n + 720 = 0

or n2 - 25n + 144 =0

or (n - 9)(n - 16) = 0

or n = 9, 16

But n = 16 is not possible as it gives the last term of A.P.

= a + (n - 1)d = 120 + (16-9) x 5

= 195° Thus the number of sides = 9.

33. Find the 20th and nth terms of the G.P.

• Solution:

G.P is

Here, a = r = =

Tn = arn – 1

Thus T20 = = =

and Tn =

34. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

• Solution: r = 2







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a8 = ar7

192 = a × 27

192/27 = a

a12 = ar11 = = 192 × 24 = 192 × 16 = 3072

35. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 =

ps.

• Solution: Let a and r be the first term and common ratio respectively. 5th term = a5 = ar

4

p = ar4

a8 = ar7

q = ar7

a11 = ar10

s = ar10

q2 = (ar7)2 = a2 r14

ps = ar4 ar10 = a2r4+10 = a2 r14

∴ q2 = ps

36. The 4th term of a G.P. is square of its 2nd term, and the first term is –3.

Determine its 7th term.

• Solution: Let a and r be the first term and common ratio respectively. Then a = -3

4th term = a4 = ar3 = -3r3

2th term = a2 = ar = -3r a4 = (a2)

2

-3r3 = (-3r)2

-3r3 = 9r2







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r = -3 7th term = a7 = ar

6 = -3(-3)6 = -3(-3)6= (-3)7 = - 2187

37. Which term of the geometric sequence:

(a) a = , r = 3/ =

(b) a = , r =

• Solution:

(a) a = , r = 3/ = Let 729 be the nth term of the G.P.

an = arn-1

729 = ( )n-1

( )12 = ( )n

n = 12 Therefore 729 is the 12th term of the G.P.

(b) a = , r =

Let be the nth term of the G.P.

an = arn-1

=

= n = 9 Therefore 1/19683 is the 9th term of the G.P.

38. For what values of x, the numbers , x, are in G.P?







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• Solution:

x2 = × = 1

x = ± 1

Therefore the required values of x are ± 1.

39. Find the sum of indicated terms of the following geometric progressions.

0.15, 0.015, 0.0015, …..; 20 terms.

• Solution: a = 0.15, r = 0.015/0.15 = 0.1 < 1

Therefore Sn = = = = = Substituting n = 20

S20 =

40. Find the sum of indicated terms of the following geometric progressions.

, , 3 , ….; n terms.

• Solution:

a = , r = > 1

Therefore Sn =

=

=

=

=







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41. Find the sum of indicated terms of the following geometric progressions. 1, -a, a2, -a3, ….; n terms. (a ≠≠≠≠ -1)

• Solution: a = 1, r = -a < 1

Therefore Sn = = =

42. Find the sum of indicated terms of the following geometric progressions. x3,

x5, x7, ….; n terms. (x ≠ ±1)

• Solution: a = x3, r = x2 < 1

Therefore Sn = =

43. Evaluate .

• Solution:

= (2 + 3) + (2 + 32) + (2 + 33) + ….. + (2 + 311).

= (2 + 2 + ………….. 11 times) + 3 + 32 + 33 + ….. + 311. a = 3, r = 32/3 = 3 > 1

Sn = = = Substituting n = 11

S11 =

Therefore the required sum is 22 + .

44. The sum of first three terms of a G.P. is and their product is 1. Find the common ratio and the terms.

• Solution: Let three terms in G.P. are







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Their product = = 1

or a3 = 1

or a = 1

Again =

or =

or = r

or 1 + 10r + 10r2 = 39r

or 10r2 - 29r + 10

or 10r2 - 25r – 4r + 10 = 0 or 5r(2r – 5) – 2(2r – 5) = 0 or (2r – 5) (5r – 5) = 0

or r =

Case (i) When r =

Numbers are

Case (ii) When r =

Numbers are

or or

Hence three numbers are or







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45. How many terms of the G.P. 3, 32, 33, …… are needed to give the sum 120?

• Solution: Given : 3 + 32 + 33+ ….. = 120

Now a = 3, r = 3. Let number of terms = n

Also = 120

or = 120

or 3n – 1 =

or 3n – 1 = 80

or 3n = 81 = 34

Hence n = 4

46. The sum of the first three terms of a G.P is 16 and the sum of the next three

terms is 128. Determine the first term, common ratio and the sum to n terms of the G.P.

• Solution: Let the three terms be a, ar, ar2.

a + ar + ar2 = 16 ………………………(i)

Let the next three terms be ar3, ar4, ar5.

ar3 + ar4 + ar5 = 128

r3(a + ar + ar2) = 128

r3(16) = 128 (from i)

r3 = 8

r = 2

Therefore the common ratio is 2.

Substituting r = 2 in (i)

a + 2a + 4a = 16







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7a = 16

a = 16/7

Sn =

Therefore the first term is 16/7, common ratio is 2 and sum to n terms is .

47. Given a G.P. with a = 729 and 7th term 64, determine S7.

• Solution: a = 729 7th term = 64 ∴ a7 = ar

6, where r is the common ratio.

729r6 = 64

r6 = 64/729

r6 = (2/3)6

r = 2/3 < 1

S7 =

=

=

=

=

= = 2059







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∴ S7 = 2059

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

the third term.

• Solution: a + ar = -4 a(1 + r)= -4 ………..(1) ar4 = 4ar2

r2 = 4 r = ±2 Substituting r = 2 in (1) a(1+2) = -4

a = Substituting r = -2 in (2) a(1-2) = -4 a = 4

... If a = ,r=2 or a = 4, r = -2.

49. 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.

• Solution: a4 = ar

3

x = ar3 …………………..(i)

a10 = ar9

y = ar9…………………..(ii)

a16 = ar15







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z = ar15…………………..(iii)

(ii) ÷ (i) y/x = ar9/ar3 = r6 …………………(iv)

(iii) ÷ (ii) z/y = ar15/ar9 = r6 …………………(v) From (iv) and (v)

y/x = z/y y2 = xz.

∴ x, y, z are in G.P

50. Find the sum to n terms of the sequence 8,88,888,888,…..

• Solution:

Sn = 8 + 88 + 888 + 8888 + …… n terms

Sn = 8[1 +11 +111 + 1111 +……. n terms]

=

=

=

=

=

51. Find the sum of the products of the corresponding terms of the sequence 2, 4, 8,

16, 32 and 128, 32, 8, 2, .

• Solution: First sequence : 2, 4, 8, 16, 32

Second sequence : 128, 32, 8, 2, ½







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New sequence after finding the products of the corresponding terms

2 × 128, 4 × 32, 8 × 8, 16 × 2, 32 ×

or 256, 128, 64, 32, 16

Now a = 256 r = = n = 5

Now Sn =

Sn =

=

= 512 × = 496

52. 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.

• Solution: Ist G.P :a, ar, ar2, …….arn – 1

2nd G.P. : A, AR, AR2,….. ARn – 1

The sequence formed after multiplying the corresponding terms of the sequences is

(aA), (aA),(rR), (aR)r2R2,…..(aA)rn – 1 Rn – 1

Here = = rR

= = rR







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= = rR

Since the ratios of two succeeding terms are the same, the resulting sequence is also a G.P.

The common ratio of the new G.P. = (rR).

53. Find four numbers forming a Geometric progression in which the third term is

greater than the first by 9, and the second term is greater than the fourth by 18.

• Solution: The four terms of G.P are a, ar, ar2, ar3. ar2 - a = 9 a (r2-1) = 9 ………..(1) ar (1 - r2) = 18 ……..(2) Dividing (2) by (1)

r = -2 Substitute r in (1) a (4-1) = 9 a = 3 ... the four numbers are 3, 3(-2), 3(-2)2 and 3(-2)3

(i.e) 3, -6, 12, -24.

54. If the pth, qth and rth terms of a G.P. are a, b, c respectively, prove that aq-r . br-

p . cp-q = 1.

• Solution: Let x be the first term and R the common ratio. Tp = x R

p-1 = a ………(1) Tq = x R

q-1 = b ……….(2) Tr = x R

r-1 = c ………(3)







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aq-r = = ...........(1)

br-p = ………….(2)

cp-q = ……….(3) L.H.S. = (1) x (2) x (3)

. =1.

55. If the first and the nth terms of a G.P are a and b respectively and if p is the

product of the first n terms, prove that P2 = (ab)n.

• Solution: Let the G.P be c, cr, cr2, ……….crn

1st term c = a nth term cr n-1 = b If n terms is c, cr, cr2, ……….crn-1 Product of n term p = c.(cr)(cr2)(cr3)……..(crn-1) = cn.r1+2+………n-1

= cn

P2 = (cn)2 2 = c2nrn(n-1) = c2n rn2-n ………….(1)

(ab)n = (c.crn-1)n = (c2rn-1)n = c2n rn2-n ……….(2)

From (1) and (2), P2 = (ab)n.

56. 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 1/rn.

• Solution: Consider the G.P. a, ar, ar2,…….







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Sn =

Sum of terms from (n + 1)th to (2n)th terms

= S2n - Sn

=

Now Required ratio =

=

=

=

=

= = R.H.S.

57. If a, b, c, d are in G.P., show that (a2 + b2 + c2)(b2 + c2 + d2) = (ab + bc + cd)2

• Solution: Since a, b, c, d are in G.P b/a = c/b = d/c = r, where r is the common ratio. ∴€b = ar, c = br = ar2 and d = cr = ar3. L. H. S = (a2 + b2 + c2)(b2 + c2 + d2) = [a2 + (ar)2 + (ar2)2][( ar)2 + (ar2)2 + (ar3)2]







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= (a2 + a2r2 + a2r4 )(a2r2 + a2r4 + a2r6) = a4 r2 (1 + r2 + r4 )(1 + r2 + r4) = a4 r2 (1 + r2 + r4 )2

R.H.S = (ab + bc + cd)2

= (a× ar + ar × ar2 + ar2 ×ar3)2

= (a2r + a2r3 + a2r5)2

= (a2r)2(1 + r2 + r3)2

= a4 r2 (1 + r2 + r4 )2 L.H.S = R. H. S

58. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

• Solution: Let the two numbers between 3 and 81 be G1 and G2.

Thus 3, G1, G2, 81 are in G.P.

Let r be the common ratio.

Therefore ar3 = 81

or 3r3 = 81

or r3 = 27 or r = 3

or G1 = 3 × 3 = 9

G2 = 9 × 3 = 27

Hence the two required numbers are 9 and 27.

59. Find the value of n so that (an+1 + bn+1)/(an + bn) may be the geometric mean

between the a and b.

• Solution:

Gn =

(an+1 + bn+1)/(an + bn) =

an+1 + bn+1 = a1/2b1/2(an + bn)

an+1 + bn+1 = an + (½) b1/2 + a1/2bn + (½)







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an+1 - an + (½) b1/2 = a1/2bn + (½) - bn+1

an + (½) (a½ - b½) = bn + (½) (a½ - b½) an + (½) = bn + (½)

ana(½) = bn b(½)

(a/b)n = (a/b)-½

n = -½ Hence value of n is -½.

60. The sum of two numbers is 6 times their geometric mean. Show that the

numbers are in the ratio 3 + 2 √√√√2 : 3 – 2√√√√2.

• Solution: Let the two numbers be a and b.

Then a + b = 6

(a + b)2 = 36ab

a2 + b2 + 2ab = 36ab

61. If A and G be A.M. and G.M. respectively between two positive numbers, prove

that the numbers are A ±±±± .







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• Solution: Let the two positive numbers be a and b.

Therefore, A = ⇒ a + b = 2A

and G = ⇒ ab = G2

The equation whose roots are a and b is

x2 – (a + b)x + ab = 0

⇒€x2 – 2Ax + G2 = 0 …….(i)

or x =

= A ±

= A ±

The roots of (i) are a and b,

Hence, the two positive number are given by

A ±

62. The number of bacteria in a certain culture doubles very 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? nth hour?

• Solution: As per question, the number of bacteria in a certain culture doubles every hour. So, they form a G.P. in which a = 30 and r = 2.

Bacteria present after 2nd hour, 4th hour and nth hour are T3, T5 and Tn+1 i.e., ar2, ar4 and

arn.

i.e., 30(2)2, 30(2)4 and 30(2n) respectively.

63. What will Rs. 500 amounts to in 10 years after its deposit in a bank which

pays annual interest rate of 10% compounded annually?







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• Solution: T1 = amount at the end of first year

= 500 = 500 ×

T2 = 500 × …..

Here a = 500 × r =

n = 10

Tn = arn – 1

T10 = 500 ×

= Rs. 500

64. If A.M. and G.M. of roots of a quadratic equation are 8 and 5 respectively, then

obtain the quadratic equation.

• Solution: Let the quadratic equation be

(x - α )(x - β ) = 0

or x2 - (α + β )x + α β = 0

Roots are x = α , β

Therefore = 8 and = 5

or α + β = 8, and = 5

or = 25

Required quadratic equation is

x2 – 8x + 25 = 0







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65. Find the sum of the following series upto n terms:1 ×××× 2 + 2 ×××× 3 + 3 ×××× 4 +

4 ×××× 5+……

• Solution: Let Tn denotes the nth term of the given series

Tn = [nth term of 1, 2, 3 …..] × [nth term of 2, 3, 4 …..]

= [1 + (n – 1) 1] [2 + (n – 1) 1]

= n(n + 1)

Tn = n2 + n

Sn = ∑ n2 + ∑ n

=

=

=

=

=

=

66. Find the sum of the following series upto n terms 1 ×××× 2 ×××× 3 + 2 ×××× 3 ×××× 4 +

3 ×××× 4 ×××× 5 +…….

• Solution: Let Tn denote the nth term of the given series .

Then,

Tu = [nth term of 1, 2, 3,….]

[nth term of 2, 3, 4,……] [nth term of 3, 4, 5,……]







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= [1 + (n – 1)1] [2 + (n – 1).1] [3 + (n –1).1] = n(n + 1) (n + 2)

= n(n2 + 2n + 3) = n3 + 3n2 + 2n

∴ Sn = ∑ n3 + 3∑ n2 + 2∑ n

=

=

=

=

=

67. Find the sum of the following series upto n terms 3 ×××× 12, 5 ×××× 22, 7 ×××× 32,……..

• Solution: Let Tn donotes the nth term of the given series. Then,

Tn = [nth term of 3, 5, 7,….]

[nth term of 1, 2, 3,….]2

= [3 +(n – 1)2][1 + (n – 1).1]2

= (2n – 1) (n)2

= n2(2n – 1) = 2n3 + n2

Hence Sn = 2∑ n3 + ∑ n2

= 2.

=

=







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=

68. Find the sum of the following series upto n terms

• Solution:

Tn =

=

=

Let = ……(i)

⇒ 1 = A(n + 1) + Bn ……(ii)

[On multiplying both sides by n(n+1)]

To find A : Putting n = 0 in (ii), we get,

1 = A(0 + 1) ⇒ A = 1 To find B : Putting n = 0 in (ii), we get

1 = B(-1) ⇒ B = -1

Putting these values of A and B in (i), the partial fractions are

=

or Tn =

Putting n = 1, 2, 3,……n, we get,

T1 =







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T2 =

T3 = ………………….

Tn =

Adding vertically, we get,

Sn = 1 - =

Extra. Hence find sum to infinity:

Now, Sn = =

As n → ∞

or S∞ = = 0

69. Find the sum of the following series upto n terms 52 + 62 + 72 + ……+ 202

• Solution: Tn of given series

= (nth term of 5, 6, 7,…..)2

= [5 + (n + 1).1]2

= (n + 4)2 = n

2 + 8n + 16

Sn = ∑ n2 + 8∑ n + 16 × n

=

=







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=

=

Put n = 16, then

S =

=

=

= 8 × 355 = 2840.

70. Find the sum of the following series upto n terms 3 ×××× 8 + 6 ×××× 11 + 9 ×××× 14+….

• Solution: Here the series is formed by multiplying the corresponding terms of two series both of which are A.P.

Viz. 3, 6, 9 …… and 8, 11, 14…..

Tn of given series

= (nth term of 3, 6, 9,….) × (nth term of 8, 11, 14,….)

= [3 +(n – 1)3] [8 + (n – 1)3]

= [3n][3n + 5]

= 9n2 + 15n

Sn = 9 ∑ n2 + 15∑ n

= 9 ×

=

= = 3n(n + 1) ( n + 3)







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71. 12+ (12 + 22) + (12 + 22 + 32) + ……

• Solution:

an = 12 + 22 + … + n2 = = =

Sn =

=

=

= (n(n+1) + 2n + 1+ 1)

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

= (n2+3n+2)

= (n+1)(n+2)

=

72. Find the sum to µµµµ terms of the series in Exercise 8 to 10 whose nth terms is

given by n(n+1)(n + 4)

• Solution: Tn = n(n+1)(n + 4)

Tn = n(n2+5n+4) = n3 + 5n2 + 4n

Sn = ∑ n

3 + 5∑ n2 + 4∑ n

=

=







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=

=

=

=

73. n2 + 2n

• Solution: an = n

2 + 2n

Sn =

=

=

=

=

= + 2(2n - 1).

74. Find the sum to n terms of the following series whose nth term is given by (2n

– 1)2.

• Solution: Tn = (2n – 1)

2

Tn = 4n2 – 4n + l

Sn = 4∑ n2 - 4∑ n + n







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EduEduEduEdumassmassmassmass

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=

=

=

=

=

=







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ch - 9 sequences and series - badhan education 11th/class 11th... · 5 corresponding series = 3 +...

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