1st PUC Maths Question Bank Chapter 9 Sequences and Series

   

Students can Download Maths Chapter 9 Sequences and Series Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Maths Question Bank Chapter 9 Sequences and Series

Question 1.
Define sequence of numbers.
Answer :
Arrangement of numbers one after another, according to some rule is called a sequence of numbers. OR Sequence is a real valued function whose domain is set of all natural numbers.

Note:

  • The numbers in the sequence are called terms.
  • The nth term of the sequence is denoted by tn or Tn or an.

Question 2.
Define finite and infinite sequence.
Answer :
The sequence which has finite number of terms, then it is called finite sequence.
The sequence which has infinite number of terms, then it is called infinite sequence.

KSEEB Solutions

Problems

Question 1.
Write the first five terms of each of the sequences whose nth terms are
(i) an = n(n + 2)
(ii) \(a_{n}=\frac{n}{n+1}\)
(iii) \(a_{n}=2^{n}\)
(iv) \( a_{n}=\frac{2 n-3}{6}\)
(v) \(a_{n}=(-1)^{n-1} \cdot 5^{n+1}\)
(v)\(a_{n}=n\left(\frac{n^{2}+5}{4}\right)\)
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 1
1st PUC Maths Question Bank Chapter 9 Sequences and Series 2
1st PUC Maths Question Bank Chapter 9 Sequences and Series 3

Question 2.
Find the indicated terms in each of the following sequences whose nth terms are given
(i) \(a_{n}=4 n-3 ; a_{17}, a_{24}\)
(ii)\(a_{n}=\frac{n^{2}}{2^{n}} ; a_{7}\)
(iii)\(a_{n}=(-1)^{n-1} \cdot n^{3} ; a_{9}\)
(iv) \(a_{n}=\frac{n(n-2)}{n+3} ; a_{20}\)
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 4

KSEEB Solutions

Question 3.
Write the first five terms of the sequences and obtain the corresponding series.
(i)\(a_{1}=3, a_{n}=3 a_{n-1}+2, \text { for all } n>1\)
(ii) \(a_{1}=-1, a_{n}=\frac{a_{n-1}}{n}, n \geq 2\)
(iii)\(a_{1}=a_{2}=2, a_{n}=a_{n-1}-1, n>2\)
Answer:
(i) Given a1 = 3, an = 3an-1 + 2, for all n > 1
Substituting n = 2, 3, 4, 5 we get
a2 = 3(a1 +2 = 3(3) +2 = 11
a3 = 3(a2) + 2 = 3(11) + 2
= 33 + 2 = 35
a4 = 3(a3) + 2 = 3(35) + 2
= 105 + 2 = 107 a5
= 3(a4) + 2 = 3(107) + 2
= 321 + 2 = 323
∴ Required five terms are ax = 3,
a2= 11, a3 = 35, a4 = 107, a5 = 323
Hence, the corresponding series is
3 + 11 + 35 + 107 + 323 +………………….
1st PUC Maths Question Bank Chapter 9 Sequences and Series 5
1st PUC Maths Question Bank Chapter 9 Sequences and Series 6

Question 4.
The Fibonacci sequence is defined by 1 = a1= a2 and an=an-1+an-2, n> 2. Find
\(\frac{a_{n+1}}{a_{n}}\) for n=1,2,3,4,5.
Answer:
Given a1 =1, a2=1,an = an-1 + an-2, n> 2
1st PUC Maths Question Bank Chapter 9 Sequences and Series 7

Arithmetic Progression (A.P.):

Question 1.
Define arithmetic progression. Answer : Let a1 ,a2, a3, be a sequence.
If a2 – a1 = a3 – a2 = a4 – a3 …………  i.e., the difference between any two consecutive terms is always same constant, then the sequence is called an Arithmetic progression (AP). The constant difference a2 – or a3 – a2 etc is called common difference and it is denoted by ‘d’.

KSEEB Solutions

Question 2.
Write the formulae for finding nth term and sum to first ‘n’ term of Arithmetic progression.
Answer :
The nth term of A.P.: If the first term of an A.P. is ’a1 ’ and the common difference is ‘d’,
then the nth an = a1 + (n -1 )d
Observation:
The A.P. with first term ‘a’ and common difference d’ is
a,a + d,a + 2d,a + 3d ……………….
Sum to first ‘n’ term of A.P.:
The sum to first ‘n’ terms of an A.P. with first term ‘a1‘ and common difference ‘d’ is,
\(S_{n}=\frac{n}{2}\left[2 a_{1}+(n-1) d\right] \text { or }\)
\(S_{n}=\frac{n}{2}\left[a_{1}+a_{n}\right]\)

Question 3.
Find the sum of odd integers from 1 to 2001.
Answer :
Required sum =1 + 3 + 5 +……….. + 2001
This series is in AP with a1 = 1, d = 2 (i.e., 3-1=2) and an =2001.
1st PUC Maths Question Bank Chapter 9 Sequences and Series 8

Question 4.
Find the sum of all natural numbers lying between 100 and 1000, which are multiples
of 5.
Answer :
Since 105, 110, 115,………… 995 are the numbers lying between 100 and 1000, are multiples of 5.
∴ Required sum =105 + 110+115 + ……………….+ 995.
This series in A.P. with an =105.
d = 110-105 = 5 and an= 995.
We have,
an = a, + (n -1 )d
⇒ 995 = 105 + (n -1)(5)
⇒ 995 = 105 + 5/1-5
⇒ 895 = 5n
1st PUC Maths Question Bank Chapter 9 Sequences and Series 9
1st PUC Maths Question Bank Chapter 9 Sequences and Series 10

Question 5.
In an A.P, the first term is 2 and the sum of the first five terms is one- fourth of the sum
of next five terms. Show that 20th term is – 112.
Answer :
1st PUC Maths Question Bank Chapter 9 Sequences and Series 11

Question 6.
How many terms of A.P.\(-6, \frac{-11}{2},-5, \dots \dots \dots\) are needed to give the sum – 25?
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 12
1st PUC Maths Question Bank Chapter 9 Sequences and Series 13

KSEEB Solutions

Question 7.
In an A.P., if pth term is \(\frac{1}{q}\) and qth term is \(\frac{1}{p}\),
prove that the sum of first pq terms is \(\frac{1}{2}(p q+1)\), where p≠q.
Answer :
Let ‘a’ be the first term and ‘d’ bet he common difference of an A.P. then, nth term
1st PUC Maths Question Bank Chapter 9 Sequences and Series 14
1st PUC Maths Question Bank Chapter 9 Sequences and Series 15

Question 8.
If the sum of a certain number of terms of the A.P 25,22,19,……… is 116. Find the last term.
Answer :
Let ‘a’ be the first term and ‘d’ be the common difference of an A.P. then
1st PUC Maths Question Bank Chapter 9 Sequences and Series 16

Question 9.
Find the sum of n terms of the A.P., whose kth term is 5A: +1
Answer :
Given kth term = 5k + 1
Substituting k = 1, 2, 3,……. , we get
1st term = 5(1)+ 1 = 6
2nd term = 5(2) + 1 = 11
3rd term = 5(3) + l = 16 etc.
Series in A.P. is 6 + 11 + 16 +………..
∴ Sum of first n terms
\(=S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(\Rightarrow S_{n}=\frac{n}{2}[2(6)+(n-1)(5)]\)
\(=\frac{n}{2}[12+5 n-5]=\frac{n}{2}[7+5 n]=\frac{1}{2}\left[7 n+5 n^{2}\right]\)

Question 10.
If the sum of ‘n’ terms of an A.P. is (pn + qn2), where p and q are constant find the common difference.
Answer :
Given Sn= pn + qn2 ………………(1)
Substituting n = 1, 2, we get
S1= p(1) + q(1)2 = p + q and
S2 =p(2) + q(2)2 =2p + 4q
we know that, S1=a = first term and S2 = first term + second term = a + (a + d),
where d = C.D.
S2 = 2a + d
= 2p + 4q = 2 (p + q) + d = 2p + 2q + d
= d = 2p + 4q-2p-2q = 2q

KSEEB Solutions

Question 11.
The sums of nterms of two arithmetic progressions are in the ratio 5n + 4 : 9n+ 6, find the ratio of their 18th
Answer :
1st PUC Maths Question Bank Chapter 9 Sequences and Series 17
1st PUC Maths Question Bank Chapter 9 Sequences and Series 18

Question 12.
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.
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 19

Question 13.
Sum of the first p, q and r term of an A.P. are a, b and c respectively. Prove that
\(\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0\)
Answer:
Let A be the first term and D be the common ratio of an A.P.
Given, Sp=a, Sq=b and Sr=c
1st PUC Maths Question Bank Chapter 9 Sequences and Series 20

KSEEB Solutions

Question 14.
The ratio of the sums of m and n terms of an A.P. is m2: it2 .Show that the ratio of mth and nth term is (2m -1): (2n -1).
Answer :
Let ‘a’ be first term and ‘d’ be the common difference of an A.P. then,
1st PUC Maths Question Bank Chapter 9 Sequences and Series 21
1st PUC Maths Question Bank Chapter 9 Sequences and Series 22

Question 15.
If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.
Answer :
Given Sn =3n2 + 5n ………..(1)
and am = mth term = 164
⇒ 164 = a + (m – 1)d ………………… (2)
Substitute n = 1, 2 in (1), we get
S1= 3(1)2 + 5(1) = 3 + 5 = 8 and
S2 = 3(2)2 +5(2) = 12+ 10 = 22
We know that, S1 = first term = a and S2 = Sum of first two terms.
∴ S1= 8 = a and a + (a + d) = 22
∴ <3 = 8 and 8 + (8 + d) = 22
⇒ d = 6
Substituting the values of a and d in (2), we get 164 = 8 +(m-1)(6)
= 8 + 6m-6 = 2 + 6m
\(\Rightarrow 162=6 m \Rightarrow m=\frac{162}{6}=27\)

KSEEB Solutions

Question 16.
If \(\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}\) is the A.M between a and b then find the value of n; a ≠ b.
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 23
1st PUC Maths Question Bank Chapter 9 Sequences and Series 24

Question 17.
Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Answer :
Let required five numbers be x1, x2, x3, x4, x5
∴ 8, x1, x2, x3, x4, x5, 26 are in A.P.
⇒ First term = 8 and seventh term
= 26 ⇒ a = 8 and a + 6d-26
⇒(3 = 8 and 6d = 26 – a = 26 – 8 = 18
⇒ a = 8 and d = 3
Hence
x1= a+d = 8 + 3 = 11
x2 = a + 2d = 8 + 6 = 14
x3 = 17, x4 = 20, x5 = 23

Question 18.
Between 1 and 31, ‘m’ numbers have been inserted in such a way that the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the value of
Answer :
1st PUC Maths Question Bank Chapter 9 Sequences and Series 25
1st PUC Maths Question Bank Chapter 9 Sequences and Series 26

KSEEB Solutions

Question 19.
A man starts repaying a loan as first installment of Rs. 100. If he increases the installment by Rs. 5 every month, what amount he will pay in the 30th installment?
Answer :
By data, the installments paid by the man are
100, 105, 110, ………………  which is in A.P. with
a = 100 and d = 5
Hence 30th installment = 30th term
= a + 29d =100 + 29(5) =245

Question 20.
The difference between any two consecutive interior angles of a polygon is 5° (5 degree). If the smallest angle is 120°, find the number of the sides of polygon.
Answer :
Let the number of sides of the polygon be n, then the sum of all exterior angles = 360° and sum of all interior angles = 180° n – 360°.
Given, smallest angle = 120° and difference between the angles = 5°.
∴ Angles are of the form,
1st PUC Maths Question Bank Chapter 9 Sequences and Series 27

Geometric Progression (G.P.):

Question 1.
Define geometric progression.
Answer :
A sequence of non-zero terms is called a geometric progression if the ratio of any term to its preceding term is constant. This constant is called common ratio and is denoted by ‘r’

Question 2.
Write the general term of a G.P.
Answer :
Let a be the first term and r(≠0) be the common ratio of a G.P. Then general term of G.P. is
Tn = a . rn-1

Note:

  • Terms in G.P. are in the form a, ar, ar2, arn-1,
  • No term in G.P. can be zero.
  • Three terms a,b,c are in G.P.
    \(\text { iff } \frac{b}{a}=\frac{c}{b} \text { i.e. }b^{2}=a c, a b c \neq 0\)

KSEEB Solutions

Question 3.
Write the formula for finding the sum of n terms of a G.P.
Answer :
Let ‘a’ be the first term and ’r’(≠) be the common ratio of the given G.P. The general teams of G.P. If Sn denotes the sum of ‘n’ terms, then,
\(S_{n}=a\left(\frac{1-r^{n}}{1-r}\right) r \neq 1\)
This can also written as
\(S_{n}=a\left(\frac{r^{n}-1}{r-1}\right)\)

Note:
(1) Sn=a + ar + ar2+….. +arn-1
(2) If r = 1, then Sn = a + a + a +……………+a – na
(3) If r = 1, then formulas for Sn fail
(4) If r < 1, then use \([S_{n}=a\left(\frac{1-r^{n}}{1-r}\right)\)
(5) If r > 1, use \(S_{n}=a\left(\frac{r^{n}-1}{r-1}\right)\)
(6) Sum of an infinite G.P., a + ar + ar2 +……. is \(\frac{a}{1-r},-1<r<1\)
(7) If a, b, c are in G.P. then ‘b’ is the G.M. (geometric mean) between a and c, then b2 =ac – \(\Rightarrow b=\sqrt{a c}\)
(8) If A and G are A.M. and G.M. of two given distinct positive real numbers, then relation between A and G is A > G.

Question 4.
Find the 20th and nth terms of the G.P.:
\(\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots \dots .\)
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 28

Question 5.
Find the 12th term of a G.P. whose 8th term is 192 and common ratio is 2
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 29

KSEEB Solutions

Question 6.
The 5th, 8th and 11th terms of G.P. are p, q and s respectively. Show that q2 = ps
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 30

Question 7.
The 4th term of a G.P. is square of its second term and the first term is – 3. Determine its 7th term.
Answer:
Let G.P. be a, ar, ar2,………….. , then
Tn=arn-1
Given: a = -3 and T4 =(T2)2
1st PUC Maths Question Bank Chapter 9 Sequences and Series 31
Question 8.
Which term of the sequence
(a) \(2,2 \sqrt{2}, 4, \cdots \cdots \text { is } 128 ?\)
(b) \(\sqrt{3}, 3,3 \sqrt{3}, \cdots \cdots \cdots \text { is } 729 ?\)
(c) \(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots \dots \text { is } \frac{1}{19683} ?\)
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 32

Question 9.
For given value of x, the number \(-\frac{2}{7}, x, \frac{-7}{2}\) are in G.P.?
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 33
1st PUC Maths Question Bank Chapter 9 Sequences and Series 34

Question 10.
Find the sum of indicated terms in each of the following G.P.
(a) 15,0.015, 0.0015,……. 20 terms.
(b) \(\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \dots \dots n \text { terms. }\)
(c) 1,-a ,a2, -a3,……….. n terms (if a ≠ – 1)
(d) x3, x5, x7,………. n terms (x ≠ ± l)
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 35

1st PUC Maths Question Bank Chapter 9 Sequences and Series 36

KSEEB Solutions

Question 11.
Evaulate:
\(\sum_{k=1}^{11}\left(2+3^{k}\right)\)
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 37

Question 12.
The sum of first three terms of a G.P. is \(\frac{39}{10} \) and their product is 1. Find the common ratio and the terms.
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 38

Question 13.
How many terms of G.P. 3, 32, 33,34,……………… are needed to give the sun 120?
Answer :
1st PUC Maths Question Bank Chapter 9 Sequences and Series 39
1st PUC Maths Question Bank Chapter 9 Sequences and Series 40

KSEEB Solutions

Question 14.
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.
Answer:
Let ‘a’ be the first term and V be the common ratio of the G.P. then a, ar, ar2,…… are in G.P.
1st PUC Maths Question Bank Chapter 9 Sequences and Series 41

Question 15.
Given a G.P. with a = 729 and 7th term 64, determine S7.
Answer :
Let ‘r’ be the common ratio.
1st PUC Maths Question Bank Chapter 9 Sequences and Series 42
1st PUC Maths Question Bank Chapter 9 Sequences and Series 43

KSEEB Solutions

Question 16.
Find a G.P. for which sum of the first two terms is – 4 and fifth terms is 4 times the third term.
Answer :
Let ‘a’ bet the first term and V be the common ratio of a G.P.
1st PUC Maths Question Bank Chapter 9 Sequences and Series 44
1st PUC Maths Question Bank Chapter 9 Sequences and Series 45

Question 17.
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
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 46

KSEEB Solutions

Question 18.
Find the sum to ‘n’ terms of the sequence 8, 88,888, 8888,………
Answer :
1st PUC Maths Question Bank Chapter 9 Sequences and Series 47

Question 19.
Find the sum of the product of the corresponding terms of the sequences 2, 4, 8,
16,32 and 128,32, 8,2, \(\frac{1}{2}\)
Answer :
Given two sequences are, 2, 4, 8, 16, 32 and 128, 32, 8,2\(\frac{1}{2}\)
∴ Required sum = 2 x 128 + 4 x 32 + 8 x 8 + 16 x 2 + 32 x\(\frac{1}{2}\)
1st PUC Maths Question Bank Chapter 9 Sequences and Series 48

Question 20.
Show that the products of the corresponding terms of the sequences
a, ar, ar2 ,……….arn-1 and A, AR, AR2,……….. AR-1 form a G.P. and find the common ratio.
Answer :
Given sequences are, a, ar, ar2 ,……….arn-1……………….(1)
and A,AR,AR2,………. ,AR-1 ……………….(2)
Now, multiply the corresponding terms of equations (1) and (2), we get
aA, aArR, aAr2R2,….
i.e., (aA),(aA)(rR), (aA)(rR)2,..
which is in G.P. with common ratio rR.

KSEEB Solutions

Question 21.
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 term by 18.
Answer :
Let ‘a’ be the first term and ‘r’ be the common ratio of a G.P.            .
Four numbers in G.P. be a,ar,ar2 and ar3
Given that T3 = Tx + 9 and T2=T4+18.
i.e., ar2-a + 9 …………….. (1)  and
ar = ar3+18 ……………… (2)
From (1), we get
ar2 – a = 9
⇒ a(r2 -l) = 9   …(3)
From (2), we get
ar – ar3 =18
⇒ nr(1- r2) = 18 …(4)
Dividing (4) by (3), we get
r = 2 ⇒ r = -2
Substituting r = – 2 in (1), we get
a( 4) = a + 9
⇒ 3a = 9
⇒ a = 3 Hence, four numbers are,
⇒ 3, 3(-2), 3(-2)2, 3(-2)3 i.e., 3,-6,12,-24

Question 22.
If the first and nth term of a G.P. are a and b respectively, and if ‘P’ is the product of ‘n’ terms, prove that aq – r ,br – p ,cr – q = 1
Answer :
Let A and R be the first term and common ratio of a G.P. Then T=ARn-l.
Let A be the first term and R be the common ratio of a G.P.
1st PUC Maths Question Bank Chapter 9 Sequences and Series 49

KSEEB Solutions

Question 23.
If the first and 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
Answer :
Let A be the first term and R be the common ratio of a G.P.
1st PUC Maths Question Bank Chapter 9 Sequences and Series 50
1st PUC Maths Question Bank Chapter 9 Sequences and Series 51

Question 24.
Show that the ratio of the sum of first ‘n’ terms of a G.P. to the sum of the terms from
\((n+1)^{\mathrm{th}} \text { to }(2 n)^{\mathrm{th}} \text { term is } \frac{1}{r^{n}}\)
Answer:
Let ‘a’ be the first term and ‘r’ be the common ratio of a G.P
1st PUC Maths Question Bank Chapter 9 Sequences and Series 52

KSEEB Solutions

Question 25.
If a, b, c, d are in G.P. show that (a2 + y2 +c2)(b2 +c2 +d2) = (ab + bc +cd)2
Answer:
Given a,b,c,d are in G.P.
Let a = a, b = ar, c = ar2 and d = ar3 and common ratio ‘r’.
LHS = (a2 +b2 +c2)(b2 +c2 +d2)
= (a2+a2r2 + a2r4)(a2r2 + a2r4 +a2r6)
= a2(1 + r2 + r4)a2(r2 +r4 + r6)
= aV(1 + r2 + r4)(l + r2 + r4)
= a4r2(1 + r2+ r4)2 ……………………(1)
RHS = (ab + bc + cd)2
= (a2r + a2r3 + a2r5)2
= [a2r(1 + r2 + r4)]2
= a4r2 (1 + r2 + r4)2 ……………………(2)
From (1) and (2)

Question 26.
Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Answer :
We b and c be the numbers between 3 and 81.
a-3, b = ar = 3r, c = ar2 = 3r2 and d = ar3 = 3r3 =T4
But T4 =ar4-1 = 3r3
⇒ 81 = 3r3
⇒ r3 = 27 = 33
⇒ r = 3
∴  b = ar = 3 x 3 = 9
and c = ar2 =3 x 32 = 27

KSEEB Solutions

Question 27.
Find the value of ‘n’so that\(\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}\) may a +b be the geometric mean between a and b.
Answer :
1st PUC Maths Question Bank Chapter 9 Sequences and Series 53

Question 28.
The sum of two numbers is 6 times their geometric mean, show that the numbers are in the ratio \((3+2 \sqrt{2}):(3-2 \sqrt{2})\)
Answer:
Let the number be a and b
Where a>b>0 the given
\(a+b=6 \sqrt{a b} \Rightarrow a+b=3(2 \sqrt{a b})\)
1st PUC Maths Question Bank Chapter 9 Sequences and Series 54

KSEEB Solutions

Question 29.
If A and B be AM and GM respectively between two positive numbers, prove that the numbers are  \(A \pm \sqrt{(A+G)(A-G)}\)
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 55
1st PUC Maths Question Bank Chapter 9 Sequences and Series 56

Question 30.
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?
Answer :
Number of bacterias increasing in each hour is given by
30 x 2 x 30 x 22,30 x 23,30 x 24…….. is in G.P.
Number of bacteria present at the end of,
2nd hour = 30 x 22 = 120
4th hour = 30 x 24 = 30 x 16 = 480 and nth hour = 30 x 2n

Question 31.
What will Rs. 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Answer :
Given the rate of interest is 10% compound annually.
1st PUC Maths Question Bank Chapter 9 Sequences and Series 57

Question 32.
If AM and GM of roots of a quadratic equation are 8 and 5 respectively, then obtain the quadratic equation.
Answer:
Let m and n be the roots, then
\(\frac{m+n}{2}=A M=8 \Rightarrow m+n=16 \text { and } \sqrt{m n}=G M=5\)
Required quadratic equation is
⇒ x2 – (sum of the roots)x + product of the roots = 0
⇒ x2 -16 x + 25 = 0
Sum to ‘n’ terms of special series
1st PUC Maths Question Bank Chapter 9 Sequences and Series 58

Exercise 9.4

I. Find the sum to ‘n’ terms of each of the series.

Question 1.
1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 +………….
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 59

KSEEB Solutions

Question 2.
1 x 2 x 3 x +2 x 3 x 4 x 3 x 4 x 5+…………….
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 60
1st PUC Maths Question Bank Chapter 9 Sequences and Series 61

Question 3.
\(3 \times 1^{2}+5 \times 2^{2}+7 \times 3^{2}+\cdots \cdots\)
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 62

Question 4.
\(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\dots+\dots\)
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 63
1st PUC Maths Question Bank Chapter 9 Sequences and Series 64

Question 5.
\(5^{2}+6^{2}+7^{2}+\cdots \cdots \cdots+20^{2}\)
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 65

KSEEB Solutions

Question 6.
3 x 8+6 x 11 + 9 x 14+……………
Answer:
Tn denote the nth term of series.
Tn = (nth term of AP 3, 6, 9……… ) (nth term of AP 8,11,14………. )
1st PUC Maths Question Bank Chapter 9 Sequences and Series 66

Question 7.
\(\begin{aligned}&1^{2}+\left(1^{2}+2^{2}\right)+\left(1^{2}+2^{2}+3^{2}\right)+\cdots \cdots\\&+\left(1^{2}+2^{2}+3^{2}+\cdots \cdots+n^{2}\right)\end{aligned}\)
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 67

KSEEB Solutions

II. Find the sum to ‘n’ terms of the series, whose nth term is given by

Question 8.
n(n +1)(n + 4)
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 68

Question 9.
n2+2n
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 69

Question 10.
(2n-1)2
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 70
Miscellaneous Exercise :

Question 1.
Show that the sum of (m + nth) and (m – n)th terms of an A.P. is equal to twice the mth
Answer :
Let ‘a’ be the first term and ‘d’ bet he common difference.
1st PUC Maths Question Bank Chapter 9 Sequences and Series 71

KSEEB Solutions

Question 2.
If the sum of three numbers in A.P. be 24 and their product is 440. Find the numbers.
Answer :
Let the three numbers be, a-d,a,a + d
Given (a-d) + a + (a + d) = 24
⇒ 3a = 24 ⇒ a = 8
1st PUC Maths Question Bank Chapter 9 Sequences and Series 72

Question 3.
Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3 respectively, show that S3=3(S2-S2)
Answer :
Let ‘a’ be the first term of d be the common difference of A.P.
1st PUC Maths Question Bank Chapter 9 Sequences and Series 73

Question 4.
Find the sum of all numbers between 200 and 400 which are divisible by 7.
Answer :
The numbers between 200 and 400 which are divisible by 7 are, 203, 210, 217,…… 399.
Required sum = 203 + 217 +…….. + 399
⇒ a = 203 and Tn = 399
⇒ 399 = 203 + (n -1)7 [∵ Tn = a + (n -1)d]
⇒ 399 = 203 + 7n-7
⇒7n = 399-203 + 7
1st PUC Maths Question Bank Chapter 9 Sequences and Series 74

KSEEB Solutions

Question 5.
Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
Answer :
The integers from 1 to 100 that are divisible by
2 or 5 are 2, 4, 5, 6, 8…….. 90, 92, 94, 95, 96, 98, 100
1st PUC Maths Question Bank Chapter 9 Sequences and Series 75

Question 6.
Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.
Answer :
Two digit numbers which when divided by 4, yields 1 as remainder are
13,17, 21,…….. , 97.
∴ Required sum = 13 + 17 + 21 +…….. + 97.
⇒ which is an A.P.
a = 13,d =4,Tn =97
∴ Tn=a + (n- 1)d
97 = 13 + (n-1)4
⇒ 4n = 97-13 + 4 88
\(\Rightarrow n=\frac{88}{4}=22\)
\(\ S_{n}=\frac{22}{2}(13+97) \quad\left[\ S_{n}=\frac{n}{2}(a+l)\right]\)
\(S_{n}=11(110)=1210\)
1st PUC Maths Question Bank Chapter 9 Sequences and Series 76

KSEEB Solutions

Question 7.
If f is a function satisfying
f(x + y) = f(x) f(y) ∀ x, y ∈ N such that f (1)
\(3 \text { and } \sum_{x=1}^{n} f(x)=120\),find the value of n.
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 77

Question 8.
The sum of some terms of G.P is 315. Whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 78
1st PUC Maths Question Bank Chapter 9 Sequences and Series 79

KSEEB Solutions

Question 9.
The first term of a G.P. is 1. The sum of the third and fifth terms is 90. Find the common ratio of G.P.
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 80

Question 10.
The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an A.P. Find the numbers.
Answer:
Let the three numbers G.P. be a, ar, ar2
1st PUC Maths Question Bank Chapter 9 Sequences and Series 81

1st PUC Maths Question Bank Chapter 9 Sequences and Series 82

Question 11.
A G.P. consists of an even numbers of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.
Answer:
Let number of terms be 2n ,an even number.
∴ G.P, be a, ar, ar2 ar3,……… ar2n-2,ar2n-1
1st PUC Maths Question Bank Chapter 9 Sequences and Series 83

Question 12.
The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11. Then find the number of terms.
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 84
1st PUC Maths Question Bank Chapter 9 Sequences and Series 85

KSEEB Solutions

Question 13.
\(\text { If } \frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}(x \neq 0)\) then show that a,b,c and d are in G.P
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 86

Question 14.
Let S be the sum, P be the product and R, the sum of reciprocals of ‘n’ terms in a G.P. prove that P2 Rn = Sn.
Answer :
Let ‘a’ be the first term and V be the common ratio of G.P
1st PUC Maths Question Bank Chapter 9 Sequences and Series 87
1st PUC Maths Question Bank Chapter 9 Sequences and Series 88

Question 15.
The pth, qth and rth terms of an A.P are a, b,c respectively . Show that
(q – r) a+(r – p)b+(p -q) c = 0
Answer:
Let A,A + D,A + 2D,…….. are in A.P.
Given pth term =A + (p – 1 )D = a ……………. (1)
qth term = A + (q – 1)D = b …………………. (2)
and rth term = A + (r-1)D = C …………………(3)
X (q- r) ⇒ a(q – r) = (q – r) [A + (p – 1)D]
X (r – p) ⇒ b(r-p) = (r-p)[A + (q – 1)D]
x(p-q) ⇒ c(p-q) = (p-q)[A + (r – 1)D]
By adding we get
⇒ a(q -r) + b(r – p) + c(p – q)
= A[(q-r) + (r-p) + (p-q) + D]
[(p -1 )(q – r) + (q – 1 )(r – p) + (r -1 )(p – q)]
= A(0) + D(0)
= 0

Question 16.
If \(a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)\) are in A.P
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 89

Question 17.
If a, b, c, d are in G.P. prove that (an + bn), (bn + cn), (cn + dn) are in G.P.
Answer:
Given a,b,c,d are in G.P.
1st PUC Maths Question Bank Chapter 9 Sequences and Series 90
1st PUC Maths Question Bank Chapter 9 Sequences and Series 91

KSEEB Solutions

Question 18.
If a and b are the roots of x2 – 3x + p = 0 and c and d are roots of x2 – 12x + q = 0.
Where a, b, c, d form a G.P. prove that (q + p) : (q-p) = 17 : 15
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 92

Question 19.
The ratio of the AM and GM of two positive numbers a and b is m : n. Show that
\( a: b=(m+\sqrt{m^{2}-n^{2}}):(m-\sqrt{m^{2}-n^{2}})\)
Answer:
We have \(A M=\frac{a+b}{2}\) and \(G M=\sqrt{a b}\)
Given Am : GM = m : n
1st PUC Maths Question Bank Chapter 9 Sequences and Series 93

Question 20.
If a, b, c are in A.P. b, c, d are in G.P. and \(\frac{1}{c}, \frac{1}{d}, \frac{1}{e}\)
are in A.P .Prove that a,c,e are in G.P.
Answer
1st PUC Maths Question Bank Chapter 9 Sequences and Series 94
1st PUC Maths Question Bank Chapter 9 Sequences and Series 95

KSEEB Solutions

Question 21.
Find the sum of the following series upto n terms.
(1) 5 + 55 + 555 +……….
(2) 0.6 + 0.66 + 0.666 +……………….
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 96
1st PUC Maths Question Bank Chapter 9 Sequences and Series 97

Question 22.
Find the 20th term of the series 2 x 4 + 4 x 6 +6 x 8+……….. + n terms.
Answer :
We have Tn = a + (n – 1)d
20th term of series
= [20th term of 2,4, 6….] x [20th term of 4, 6, 8….. ]
= [2 + (20 -1)2] x [4+ (20-1)2]
= (2+ 38)(4 + 38)
= (40)(42)
= 1680

Question 23.
Find the sum of the first ‘n’ terms of the series, 3 + 7 + 13 + 21 + 31 +…………
Answer :
Let Tn= nth term of the series and Sn = Sum of first ‘n’
1st PUC Maths Question Bank Chapter 9 Sequences and Series 98
1st PUC Maths Question Bank Chapter 9 Sequences and Series 99

KSEEB Solutions

Question 24.
If S1, S2, S3 are the sum of first ‘n’ natural numbers, their squares and their cubes, respectively, show that 9S2 = S3(1+ 8S1)
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 100

Question 25.
Find the sum of the following series up to n term
\(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\cdots\)
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 101

KSEEB Solutions

Question 26.
Show that
\(\frac{1 \times 2^{2}+2 \times 3^{2}+\cdots+n(n+1)^{2}}{1^{2}+2+2^{2} \times 3+\cdots+n^{2}(n+1)}=\frac{3 n+5}{3 n+1}\)
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 102
1st PUC Maths Question Bank Chapter 9 Sequences and Series 103

Question 27.
A farmer buys a used tractor for Rs. 12,000. He pays Rs. 6000 cash and agrees to pay the balance in annual instalments of Rs. 500 plus 12% interest on the unpaid amount. How much will the tractor cash him?
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 104

KSEEB Solutions

Question 28.
Shamshad Ali buys a scooter for Rs. 22000. He pays Rs. 4000 cash and agrees to pay the balance in annual installment of Rs. 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?
Answer :
1st PUC Maths Question Bank Chapter 9 Sequences and Series 105

Question 29.
A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.
Answer :
1st PUC Maths Question Bank Chapter 9 Sequences and Series 106
1st PUC Maths Question Bank Chapter 9 Sequences and Series 107

Question 30.
A man deposited Rs. 10,000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.
Answer:
1st PUC Maths Question Bank Chapter 9 Sequences and Series 108

Question 31.
A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
Answer:
Cost of the machine at the end of one year [larex]5625\left(1-\frac{20}{100}\right)[/latex]
1st PUC Maths Question Bank Chapter 9 Sequences and Series 109
1st PUC Maths Question Bank Chapter 9 Sequences and Series 110

KSEEB Solutions

Question 32.
150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
Answer :
Let V be the number of days that the work was to be completed
∴ For one worker, number of days required to complete the work = 150 n.
150 + 146 + 142 + ….upto (n + 8) terms = 150 n.
1st PUC Maths Question Bank Chapter 9 Sequences and Series 111
1st PUC Maths Question Bank Chapter 9 Sequences and Series 112

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