Sequences and Series Class 11

Ex 8.1, 1 teachoo
Write the first five terms of the sequences whose n" term is
a, =n(n+ 2)
We need to find first five terms
L€. a4, A, 3 ag, As
a, = n(n + 2)
Putting n = 1ina,
a, =1(1+ 2)
=1x3
=3

Ex 8.1, 2 teachoo
Write the first five terms of the sequences whose n" term is
_ n
An = n+1
We need to find first five terms
L€. a4, A, a3 ag, As
n
i =— {1
Given a, = — > (1)
Putting n = 1 in (1)
_i1
a 4d
_1
a=

Ex 8.1, 3 teachoo
Write the first five terms of the sequences whose n" term is a, = 2"
We need to find first five terms
i€, a4, a, a3 ay, as
Given a, = 2" (1)
Putting n = 1 in (1)

a, = 2%

=2

Putting n = 2 in (1)

a, = 2?

aj=2x2=4

Ex 8.1, 4 teachoo
; . . . 2n—-—3
Write the first five terms of sequences whose n" term is a, = a
We need to find first five terms
L@. a4, A, a3 ag, as
2n-3
a= 282.)
6
Putting n = 1 in (1)
_2x1-3
a, = -
_2-3
” 6
_71
~ 6

Ex8.1,5 teachoo
Write the first five terms of the sequences whose n" term is
a, = (-1)9-1 pitt
We need to find first five terms
1. ay, Ay, Az Ag, As
Let a, = (-1)r-15"+2 (1)
Putting n = 1in (1)
a, =(-1)t-151+1
= (-1)° x 52
=1x5?
=5x5=25

Ex 8.1, 6 teachoo
Write the first five terms of the sequences whose n" term is
nts
a
We need to find the first five terms
i.€. ay, a), az a, and a,
, n+
Given a, =n C=) .(1)
4
Putting n = 1in (1)
+5
a, -1 245)
4
-1 (+5) 6 3
~ 4 4 2

Ex 8.1, 7 teachoo
Find the 17" and 24" term in the following sequence whose n'”
term is a, = 4n - 3;
We need to find a,, and a5,
a, = 4n-3 (1)
Putting n = 17 in (1)
a47=4x17-3
=68-3
=65
Putting n = 24 is (1)
a4=4x 24-3

Ex8.1,8 teachoo
n?
Find the 7" term in the following sequence whose n" term is a, = ory
We need to find a,
n2
ay = mn
putting n =7
72
a;= oa
_ 7X7
"2xX2X2x2x2x2x2
_ 49
~ 128
Hence the 7" term is ae
128

Ex 8.1, 9 teachoo
Find the 9" term in the following sequence whose n" term is
a, = (1)? n?
We need to find a,
a, = (-1)"" 1 n? (1)
putting n = 9 in (1)
ay =(-1)9-* 9°
= (-1)° (9)°
=1x9?
= (9)°
= 729
Hence the 9" term is 729

Ex 8.1, 10 teachoo
Find the 20" term in the following sequence whose n" term is
_ n(@- 2)
an = n+3
We need to find ay5
n(n — 2)
a, = ———
n+3
Putting n = 20
_ 20(20 — 2)
920-3043
_ 20(18)
"23
_ 360
~ 23
_ 360
Hence the 20" term is 7

Ex 8.1, 11 teachoo
Write the first five terms of the following sequence and obtain the
corresponding series:
a, = 3,a, = 3a,_,+2 foralln>1
It is Given that
a,=3
For a, onwards, we have to use formula
a, = 3a,_1,+ 2 (1)
Putting n = 2 in (1)
a,=3a,,+2
=3a,+2 (a, = 3}
=3x3+2=9+2=11

Ex 8.1, 12 teachoo
Write the first five terms of the following sequence and obtain the
corresponding series:
a,=-1, a,==4,n22
It is given that
a,=-1, (1)
For a, and onward, we have to use this formula
a= an-1
" n
Putting n = 2 in (1)
an a2-1
a 2
-H_71 ee
=Fes> (a=-t)

Ex 8.1, 13 teachoo
Write the first five terms of the following sequence and obtain the
corresponding series:
a, =a, =2,a,=a,,-1,n>2
It is given that

a,=2

a=2
For a, onwards, we have to use formula

an =a,1-1 wa (1)
Putting n = 3 in (1)

a,=a,,-1

a3;=a,-1

=2-1=1 (a, = 2)

Ex 8.1, 14 teachoo
The Fibonacci sequence is defined by
1l=a,=a,anda,=a,,+a,)n>2.
Find mt, forn=1,2,3,4,5
Lets first calculate a,, a, a3, a4, as & ag
It is given that

a,=1

a,=1
For a3, ay, a, & a; we need to use

a, = Aya t+ ang, N>2 (1)

Ex 8.2, 1 teachoo
. th th 555
Find the 20" and n terms of the G.P. Bagh gre
_5 5 5

G.P. is PY ru a
We know that

a,=ar-?
where a, = n'" term of GP

nis the number of terms

a is the first term

ris the common ratio

; 5

Here, First term a = 3

Ex 8.2, 2 teachoo
Find the 12" term of a G.P. whose 8" term is 192 & the common
ratio is 2.
We know that

a, =ar"-t
where a, =n" term of GP

nis the number of terms

a is the first term

ris the common ratio
Here,
common ratio r = 2
& 8 term is 192

i.e. ag = 192

Ex 8.2, 3 teachoo
The 5", 8 and 11" terms of a G.P. are p, q and s, respectively.
Show that q? = ps.
We know that
a,=ar"-t
where a, = n'" term of GP
nis the number of terms
ais the first term
ris the common ratio
Here,
5" term is p
ie.a,=p (1)

Ex 8.2, 4 teachoo
The 4'" term of a G.P. is square of its second term, and the first
term is -3. Determine its 7 term.
We know that

a,=ar"-t
where a, = n'" term of GP

nis the number of terms

a is the first term

ris the common ratio
Here,
First term =a

= 33

Ex 8.2, 5 teachoo
Which term of the following sequences:
(a) 2, 2V2, 4....... is 128
2, 2V2, 4,.....
We know that
a, =ar-+
where a, = n'" term of GP
nis the number of terms
ais the first term
ris the common ratio

Which term of the following sequences:
(b) (3,) 3, 3 3, is 729 ?

Which term of the following sequences:
(c) 1/3, 1/9, 1/27, .. Is 1/19683 ?

Ex 8.2, 6 teachoo
-2 -7 .
For what values of x, the numbers Z% > arein G.P?
_ -2 =7 .
Since —, x, = are in GP
7 2
So common ratio will be same
. Second term
Common ratio (r) = ——_—___.
First term
x
2
7
7x
=— AL
= (1)

Ex 8.2, 7 teachoo
Find the sum to 20 terms in the geometric progression
0.15, 0.015, 0.0015 ...
0.15, 0.015, 0.0015 ...
We know that
_ ad - rt)

Sn = 1-r
where S,, = sum of n terms of GP

nis the number of terms

a is the first term

ris the common ratio
First term a = 0.15,

Ex 8.2, 8 teachoo
Find the sum to n terms in the geometric progression V7 ,V21, 3v7...
V7 N21 ,3V7...
Here,
First term a= V7
c tior 222
‘common ratio r = 7
_v7 x3
"7
_V7 xv3
"7
=V3

Ex 8.2, 9 teachoo
Find the sum to n terms in the geometric progression
1, -a, a7, -a°..... (ifa #-1)
1, -a, a’, -—a°.....
Since small ‘a’ is used here, we used A’ for first term
We know that
A(1—R”
5, = ¢ )
1-R

where S, = sum of n terms of GP

nis the number of terms

Ais the first term

Ris the common ratio

Ex 8.2, 10 teachoo
Find the sum to n terms in the geometric progression
x?, X°, x7... (ifx + 1)
90, x7.
We know that
_ yn

where S,, = sum of n terms of GP

nis the number of terms

a is the first term

ris the common ratio

Ex 8.2, 11 teachoo
11
Evaluate Ye + 3*)
k=1
Introduction
(2+ 3)
Atk=1, 2+31
Atk=2, 2+ 37
wae 1
Ye +34)
on rst
Atk=11, 2+ 3"
11
Ve + 3k) =(2+31)+(24+32)4+(2+39)+..4+(24+ 31)
k=1

Ex 8.2, 12 teachoo
. _ 39 . .
The sum of first three terms of a G.P. is To and their product is 1.
Find the common ratio and the terms.
1st term of G.P = <
2st term of G.P=a
3st term of G.P = ar
It is given that
' 39

sum of first three terms = To
. a tatar= 39
Le. tatar=70 (1)

Ex 8.2, 13 teachoo
How many terms of G.P. 3, 32, 3%, ... are needed to give the sum 120?
First term =a = 3
Common difference r = =

=3
We know that

ris
-1

Sum of n terms is = a" 4)

r-1
We need to find number of terms required to give sum 120
Let sum of n terms of this G.P. = 120

Ex 8.2, 14 teachoo
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.
Given
Sum of first three terms = 16 Sum of next three terms = 128
S; = 16 Sum of 4t, 5 & 6t term = 128
(Sum of first six term — Sum of first
three term) = 128
S,—S3= 128
Putting S;=16
S,—16 = 128
S,= 128+ 16
S,= 144

Ex 8.2, 15 teachoo
Given a G.P. with a = 729 and 7" term 64, determine S,.
First term a = 729
and 7" term = 64
we know that
n'* term of G.P. = ar™*
a, = ar®
Putting values
64 = 729 r®
64
729

Ex 8.2, 16 (Method 1) teachoo
Find a G.P. for which sum of the first two terms is —4 and the fifth
term is 4 times the third term.
Let a be the first term
& r be the common ratio of G.P.
It is given that
Sum of first two term is = -4
ie.5,=-4 (1)
Also
fifth term is 4 times of third term
ie.a,=4%x a;

Ex 8.2, 17 teachoo
If the 4, 10" and 16" terms of a G.P. are x, y and z, respectively.
Prove that x, y, z are in G.P.
We know that
a,=ar"-t
where a, = n"" term of GP
nis the number of terms
a is the first term
ris the common ratio
Here,
4" term is x
i.e. a, =X

Ex 8.2, 18 teachoo
Find the sum to n terms of the sequence, 8, 88, 888, 8888...
8, 88, 888, 8888... ton term
This is not a GP but we can relate it to a GP
By writing as
Sum = 8 + 88 + 888 + 8888 + ... upto n terms
= 8(1) + 8(11) + 8(111) + ... upto n term
Taking 8 common
= 8(1+ 11+ 111+ .. upton term)
Divide & multiply by 9
= sl9(4 + 11+ 111+ .. upton term)]
=5 [9 +99 + 999 + 9999 +...upton terms]

Ex 8.2, 19 teachoo
Find the sum of the products of the corresponding terms of the
sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, ;
1 sequence is 2, 4, 8, 16, 32
2"4 sequence is 128, 32, 8, 2,5
1 sequence | 2"! sequence Product of
corresponding terms
1*t term 2 128 2x 128
+
2™ term 4 32 4x 32
+
d
3 term 8 8 8x8 Sum
+
4" term 16 2 16x2
1 + 1
5" term 32 = 32x=
2 2

Ex 8.2, 20 teachoo
Show that the products of the corresponding terms of the sequences
a, ar, ar?, ar’... ar"-1 & A, AR, AR2, AR®...... AR"-' form a G.P, and find
the common ratio.
1 sequence is a, ar, ar’, ar?... ar"~*
2"4 sequence is A, AR, AR*, AR®...... AR"~ +
15°* sequence | 1°t sequence Product of
corresponding terms
1 term a a axA
+
2™ term ar AR ar x AR
+
sum
34 term ar? AR? ar? x AR?
+
n® term are ARP ar™t x ARP

Ex 8.2, 21 teachoo
Find four numbers forming a geometric progression in which third
term is greater than the first term by 9, and the second term is
greater than the 4th by 18.
Let a be the first term of G.P
r be the common ratio of the G.P.
We know that
n term of G.P = ar"
ie.a,=arn
It is given that
Third term is greater than the 1st term by 9
ie. aj=a,+9

Ex 8.2, 22 teachoo
If the p,q" and r” terms of a G.P. are a, b and c, respectively.
Prove that at" b""%c?-9 =1
We know that
n™ term of G.P =AR"-1
(We are using a, rin the question,
so we use A for first term and R for common ratio)
It is given that
p" term of G.P=a
Ap=a
ARPt=a
a= ARPT

Ex 8.2, 23 teachoo
If the first and the nth term of a G.P. are a ad b, respectively, and if
P is the product of n terms, prove that P? = (ab)".
Let a be the first term of G.P
& r be the common ratio of G.P
Given,
first term of G.P=a
We know that
n® term of G.P = ar
b=ar"t (1)

Ex 8.2, 24 teachoo
Show that the ratio of the sum of first n terms of a G.P. to the sum
of terms from (n+1)'* to (2n)* term is =
Sum of first n terms Sum of terms from (n+1)" to 2n™
term
Sum of n terms of GP is
Sum from (n+1)" to 2n” term
Sa = —_ (1) = Sum from 15 to 2n term
-r
—Sum from 1* ton term
= Sum of first 2n terms
— Sum of first n terms
= Son Sp
_ a(i-r?") _ a(i- r?)
~ d= (-r)
__ 4 _ y2ny _ _ yn
==") ==")
__4 n_ »2n
=Gun lt rr) (2)

Ex 8.2, 25 teachoo
If a, b, c and d are in G.P. show that .
(a2 + b?+ c?) (b?+ c+ d?) = (ab + bc + cd)?
We know that a, ar, ar’, ar®, .... are in G.P.
with first term a
& common ratio r
Given a, b, c, d are in G.P.
So, a=a
b=ar
c=ar*
d=ar°

Ex 8.2, 26 teachoo
Insert two numbers between 3 and 81 so that the resulting
sequence is G.P.
We know that to insert n numbers between a & b
1

b\w
Common ratio r = Gyn
We need to insert 2 numbers between 3 & 81
Herea=3&b=81
& n = number of terms inserted = 2

Ex 8.2, 27 teachoo
at +1 + br +1
Find the value of nso that ——_—-—— may be the geometric
an +pn y 8
mean between a and b.
We know that geometric mean between a & b
isa&b= vab
It is given that
at +1 + pr +41
G.M. between a & b = ———__—_
a? +p”
at +1 + pr +41
ab = ar +p”
1 at +1 + pr +1
(ab) 2 = —_—__—_
a? +p”

Ex 8.2, 28 teachoo
The sum of two numbers is 6 times their geometric mean, show
that numbers are in the ratio (3 + 2-¥2) : (3-2 V2)
Introduction
Componendo dividendo
x a
if yb
Applying componendo dividendo
x+y _a+b
x-y “a-b
Eg: Taking~=~
g: Taking > ==

Ex 8.2, 29 teachoo
If Aand G be A.M. and G.M., respectively between two positive
numbers, prove that the numbers are A+ (A +G)(A—G)
Let a & b be two numbers
We need to show that the numbers are A+ ./ (A + G)(A — G)
iLe.a=A+J(A+G)(A—-G)

b=A-J(A+G)(A-G)
Now we know that

atb

Arithmetic mean = A = >
Geometric mean = G = Vab

Ex 8.2, 30 teachoo
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 2"¢ hour, 4" hour and n" hour?
Number of bacteria | Term of G.P.
Existing 30 a,
After 1* hour 30x2 =60 a,
After 2™ hour 60x 2 =120 a;
After 3" hour 120 x2 =240 aq
Our sequence is 30, 60, 120, ....
This is a G.P. as
60 120
—=2 & —=2
30 60
Common ratio = 2
First term = 30

Ex 9,2, 31 (Method 1) teachoo
What will Rs 500 amounts to in 10 years after its deposit in a bank
which pays annual interest rate of 10% compounded annually?
We use the formula
t
r
A=P(1+ =)

Here, P = principal = 500

r= rate of interest = 10%

t=time = 10 years

A be the amount

Ex 8.2, 32 teachoo
If A.M. and G.M. of roots of a quadratic equation are 8 and 5,
respectively, then obtain the quadratic equation.
Introduction
If 2, 3 are the roots or a quadratic equation,
The quadratic equation is

x? -(2+3)x+(2x3)=0

x*-5x+6=0
Therefore,
If a & B be the root of the quadratic equation
So, the quadratic equation becomes

x? — (Sum of roots)x + (product of roots) = 0

ie. x*-(a@+ B)x+ ap =0

Example 1 teachoo
Write the first three terms in each of the following sequences
defined by the following:
(i) a,=2n+5
We need to find first three terms
i.e. ay, az, ag
Given a,=2n+5__...(1)
Putting n = 1 in (1) Putting n = 2 in (1) Putting n = 3 in (1)
a, =2(1)+5 a, = 2(2) +5 a, = 2(3)+5
F245 =4+5 =64+5
=7
=9 =11

Example 1
Write the first three terms in each of the following sequences defined by the following:
(ii) an = (𝑛 − 3)/4

Example 2 teachoo
What is the 20th term of the sequence defined by
a, =(n- 1) (2-n) (3 +n)?
We need to find 20" term i.e. azo
Given a, = (n—1) (2-1) (3 +n) ..(4)
Putting n = 20 in (1)
Ayo = (20 — 1) (2— 20) (3 + 20)

= (19) x (- 18) x (23)

=— 7866
Hence, the 20" term is -7866

Example 3 teachoo
Let the sequence an be defined as follows:
a,=1,a,=a,_,+2forn22.
Find first five terms and write corresponding series.
Given- that

a, =1,
For a, and onward
We use this formula.

a,=a,_,+2forn2>2_ ...(1)
Putting n = 2 in (1)

a,=a,_,+2

Example 4 teachoo
Find the 10th and nth terms of the G.P. 5, 25, 125,....
5, 25,125,...
We know that
a, =ar"-?
where a, = n'" term of GP
nis the number of terms
a is the first term
ris the common ratio
Here, first terma=5,
common ratio r = 2
=5

Example 5 teachoo
Which term of the G.P., 2,8,32, ... up to n terms is 131072?
Given G.P., 2,8,32, ...upto n terms
We know that
a, =ar"-t
where a, =n'" term of GP
nis the number of terms
a is the first term
ris the common ratio
Here,
First term a = 2

Example 6 teachoo
Ina G.P., the 3% term is 24 and the 6" term is 192. Find the 10"
term.
We know that

a, =ar"-?
where a, = n'" term of GP

nis the number of terms

ais the first term

ris the common ratio
Here,
3 term is 24

ie. a, = 24

Example 7 teachoo
Find the sum of first n terms and the sum of first 5 terms of the
geometric series
2 4
Ttotrot.
3.9
2 4
Ttatr ti.
3.9
We know that
a(1—r™
1-r
where S,, = sum of n terms of GP
nis the number of terms
a is the first term
ris the common ratio

Example 8 teachoo
How many terms of the G.P. 3, 5, yyy are needed to give the
3069
a
sum = *
Here
First term = a = 3,
3
Common ratio r = 3 (1)
_ 3
“2x3
1
“2

Example 9 teachoo
The sum of first three terms of a G.P. is ae and their product is — 1.
Find the common ratio and the terms.
Let the three terms in G.P. be *, a, ar
Here,

1st term of G.P. = <

2nd term of G.P.=a

3rd term of G.P. = ar
It is given that

13

Sum of first three terms of G.P. = Db

Example 10 teachoo
Find the sum of the sequence 7, 77, 777, 7777, ...ton terms.
7,77, 777, 7777, ... terms
77
Here, — =11
7
777
& — = 10.09
77
Secondterm Thirdterm |
Thus, —————— # —————— i.e. common ratio is not same
First term Second term
« This is not a GP
We need to find sum
Sum = 7+ 77+777+ 7777+ ...upton terms

Example 11 teachoo
A person has 2 parents, 4 grandparents, 8 great grandparents, and
so on. Find the number of his ancestors during the ten generations
preceding his own.
The number of ancestors are
2, 4, 8, 16, ... (upto 10" generation)
It is a G.P. since,
4
37 2
BL
4
« Common ratio = 2
First term = a=2

Example 12 teachoo
Insert three numbers between 1 and 256 so that the resulting
sequence is a G.P.
We know that to insert n numbers between a & b
1
b\a
Than common ratio r = Cy"
We need to insert 3 numbers between 1 & 256
Herea=1&b=256
—
256
Sor= ‘Gola
1
1
= (256)4

Example 13 teachoo
If A.M. and G.M. of two positive numbers a and b are 10 and 8,
respectively, find the numbers.
We know that AM of two Also GM of two number a & b is
number a & b is GM =Vab
a+b
AM ="
& given that GM of two number a
oo &bis 8
It is given that AM of two
number a &b is 10 ie. G.M.= Vab = 8
ie. AM = 10 Squaring both sides
2
a+b, (vab)’ = (8)?
2
at+b=20 (1) ab = 64 (2)

Example 14 teachoo
aoioa
If a,b, c are in G.P.and ax = bY = cz, prove that x, y, z are in A.P.
aofo2
Given that ax = bY = cz
aoioa
Let ax = bY =cz =k
Now,
2
ax =k
Taking power x both sides
41%
(ax) = (k)*
at xt ake
a=k*

toackoo.com
Example 4,
In an A.P. if mt term is n and the n' term is m, where m zn,
find the p"™ term.
We know that
a,=a+(n—1)d
ie.n term=at(n-1)d
Thus, m“ term = a,,
=at+(m-—1)}d

It is given that
m" term is n

at(m-1)d=n {1}
Also, it is given that
n® term is m

a+(n-1)d=m wa. (2)

toachoo.com
Example 5,
If the sum of n terms of an A.P. is nP + sn(n —1)Q, where
P and Q are constants, find the common difference.
Let a,, a), ... a, be the given A.P
Given, Sum of n terms = nP + ; n(n—1)Q
1

S,=nP+>n{n—1jQ AL)
Putting n = 1in (1)

S,=1xP +> x1x (1-1)Q

1

S,=P+ 319}

S,=P
But sum of first 1 terms will be the first term

aye S; =P

teachoo.com
Example 6
The sum of n terms of two arithmetic progressions are in the ratio
(3n +8): (7n +15). Find the ratio of their 12" terms.
There are 2 AP’s with different first term and common difference
For the first AP
Let first term be a
common difference be cd
Sum of nterm =S, = 5 (2a + (n—1}d)
& n term =a, =a +(n—1)d
Similarly for second AP
Let first term =A
common difference = D
S, =5(2A+ (n-1)D)
& n term =A, =A+(n-1)D

toackoo.com

Example 7
The income of a person is Rs. 3,00,000, in the first year and he
receives an increase of Rs.10,000 to his income per year for the
next 19 years. Find the total amount, he received in 20 years.
Income of the person in 1*t year = Rs 3,00,000
Income of the person in 2" year = Rs 3,00,000 + 10,000

= Rs 3,10,000
Income of the person in 3" year = Rs 3,10,000 + 10,000

= Rs 3,20,000
Thus, Income received every year is

300000, 310000, 320000...

This is an A.P as difference between consecutive terms is constant.

toackoo.com
Example 8
Insert 6 number between 3 and 24 such that the
resulting sequence is an AP
We know that to insert n number between a and b
common difference = d = 2—*
n+1
Here,
We need to insert 6 numbers between 3 and 24
So,b=24,a=3
& number of terms to be inserted = n =6
Therefore,
dq 22423
641
24
“7
=3

teachoo.com
Example 19,
Find the sum to n terms of the series:
54+114+19+29+ 41...
It is not an AP or a GP
Let
S, = 5+114+19+29+41..+a,, +a, (1)
S, = 04+54114+19+41..+a,,+a,, +4, (2)
Subtracting (2) from (1)
Sy — Sy, = 5-04 [(11 — 5) + (19 — 11) + (29-19) +...(a,_,- ay, )
+ (a) — ana) —a,
0=5+[6+8+10+12+..a,,]—-a,
a,=5+[6+8+10+12+...+(n—1) terms] (3)

teackoo.com
Example 20,
Find the sum to n terms of the series whose n" term is n (n+3).
Given
a, =n(n+ 3}
=n? +3n
The sum of n terms is
n
n=1
n
= »} n?+3n
n=1
n n
_ _ nati)
= Sines San 14#24+34..¢n=Yn=——
n=1 n=1 72422432. ¢n2 =Vnet= n(nt1)(2n+1)
n n 6
=> 43) 0 3534 28 3 3 _ (mtn?
a a 234+ 27+3 4m =n =~ )
_ n(n + 1)(@n+ 1) 4 3n(n + 1)
~ 6 2

teachoo.com
Example 21
If p®, qt, r and s terms of an A.P. are in G.P, then show
that (p—q), (q—r), (r—s) are also in G.P.
We know that the n' term of AP is a + (n— 1)d
ie.a,=a+(n—1)d
Term Value

p™ term a,=at(p—ijd

q'® term a,=at(q—ijd

term a,=at(r—l)d

s‘* term a,=a+(s—1)d
It is given that Apr Ag A, &a, in GP
i.e. their common ratio is same
S094 = 3 = Ss ...(1)

"dp 4, 4,

teachoo.com
Example 23,
If a, b, c, d and p are different real numbers such that
(a? + b? + c2)p* — 2(ab + bc + cd) p + (b* + c? +d?) < 0, then show
that a, b, c and d are in G.P.
Introduction
Ifx?+y?24+z2<0
ee
-1 -2 —3 (-1)?+(-2)4(-3)2 =14 Positive
3 2 1 (3P+(2P4(1 = 14 Positive
2 3 1 (2)4+(-3)2+(1)2 =14 Positive
0 (e) 0 (0)2+(0)2+(0)2 =O Zero
So, if x2 + y* +2? <0,
x +y24+27=0
ie.x=0,y=0,z=0

teachoo.com
Example 24
If p,q,r are in G.P. and the equations, px? + 2qx + r=0 and
dx? + 2ex + f = 0 have a commen root, then show that «, 7 7
are in A.P
Introduction
For a quadratic equation,
ax?+bx+c=0
roots are x = —

Misc 1 teackhoo
If f is a function satisfying f (x + y) = f(x) f(y) for all x, y € N such that
f(1) = 3 and y fGxo) = 120, find the value of n.
Given that :
f (x + y) = f(x) fly) Vx, y EN
and f(1) = 3
Py tte) i) toes es)
f(1+1)=f (1) xf (1)
1 1 f(1)=3 f(1)=3
f(2) =3x3=9
f (1+ 2) =f (1) x f (2)
1 2 f(1j=3 ~~ f(2)=9
f (3) =3x9= 27
2 2° f(2)=9 F(2)=9 f (2+ 2) =f (2) x f (2)
f(4)=9x9=81

Misc 2 teachoo
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.
Let a be the first term of GP
& r be the common ratio of GP
It is given that
a=5
&r=2(r>1)
Sum of some term of a GP = 315
Let the sum of n terms of GP = 315
S,= 315

Misc 3 teachoo
The first term of a G.P. is 1. The sum of the third term and fifth term
is 90. Find the common ratio of G.P.
It is given that
Sum of third term & fifth term is 90
ie. a, +a, = 90 (1)
We know that
n“ term of GP = ar®-1
ie. a, =arn-t
Puttingn=3&a=1

azeir-t

=(2

Misc 4 teachoo
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 arithmetic progression.
Find the numbers.
Let the numbers in G.P. be a, ar, and ar.
It is given that
Sum of these number is 56
atar+ar =56
ar? = 56-a-ar (1)
Also,
When 1, 7, 21 subtracting from these number respectively,
the new numbers are in AP

Misc 5 teachoo
AG.P. consists of an even number of terms. If the sum of all the
terms is 5 times the sum of terms occupying odd places, then find
its common ratio.
Introduction
Let G.P. 10, 100, 1000, 10000, 100000, 1000000
Here, number of terms = 6 which is even

100
Common ratio = —

10
=10

Sum of all terms = 10 + 100 + 1000 + 10000 + 100000 + 1000000
Sum of terms occupying odd places = 10 + 1000 + 100000

Misc 6 teachoo
if at+bx b+cx _c+dx
"a—bx b-cx a-—dx
(x # 0) then show that a, b, c and d are in G.P.
Introduction
Componendo dividendo
x a
if yb
Applying componendo dividendo
xty_at b
x-y “a-b

Misc 7 teachoo
Let S be the sum, P the product and R the sum of reciprocals of n
terms in a G.P. Prove that P?R"= S$”
Let a be the first term of GP
& rbe the common ratio of GP
We know that

ris

-1

Sum of n term of gp ==) (1)
Now, finding P

Misc 8 teachoo
If a, b, c, d are in G.P, prove that (a" + b"), (b" +c"),
(c"+ d") are in G.P.
We know that a, ar, ar’, ar®, .... are in G.P.
with first terma
& common ratio r
Given a, b, c, d are in G.P.
So,a=a
b=ar
c=ar?
d=ar
We want to show that
(a" + b"), (b" + c"), (c"+ d") are in GP

. teachoo

Misc 9
If a and b are the roots of x?- 3x+p=0 andc, dare roots of x?-
12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q—p)
= 17:15.
Introduction
For quadratic equation ax? + bx +c =0

Product of roots = <

-b
& sum of roots = 7

Misc 10 teachoo
The ratio of the A.M and G.M. of two positive numbers a and b, is
m:n. Show that a: b=(m+vm2—n?) :(m-Vm? —n? }
Introduction
Componendo dividendo
if =="
y b
Applying componendo dividendo
x+y a+b
x-y a-b
Eg:
Taking + = 4
2 8

Misc 11 teachoo
Find the sum of the following series upto n terms:
(545545554...
This is not GP but it can relate it to a GP by writing as
Sum =5+55+555 +..... ton terms

= 5(1) + 5(11) + 5(111) + ... ton terms
taking 5 common

=5(1+11+111+... ton term)
Divide & multiply by 9

= s[9(1 +11+111+... ton term)]

=5 [9+ 99+ 999 +....ton terms]

= 5 (10-1) + (100 - 1)+ (1000 - 1) + ... ton terms]

Misc 11
Find the sum of the following series upto n terms:
(ii) .6 +.66 +. 666 +…

Misc 12 teachoo
Find the 20" term of the series
2x4+4x6+6x8+...+nterms.

1%term | |2|x/4

2-4 term | |4|x16

3% term | |6|x/8

n™term | (n term of 2, 4, 6, ...) x (n term of 4, 6, 8, ...)

Misc 13 teachoo
A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash &
agrees to pay the balance in annual instalments of Rs 500 plus 12%
interest on the unpaid amount. How much will be tractor cost him?

Amount paid to buy tractor = Rs 12,000

He pays cash = Rs 6000

Remaining balance = 12000 — 6000

= 6000
Annual instalment = Rs 500 + interest@12% on unpaid amount
rane agnor inert
unpaid amount instalment

1** Instalment 6000 = = x 6000=720 =500+ 720= 1220
2"4 Instalment 6000 — 500 = 5500 = = x 5500=660 =500+ 660= 1160
34 Instalment 5500 — 500 = 5000 = = x 5000=600 =500+ 600= 1100

Misc 14 teachoo

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?

Amount paid to buy scooter = Rs 22000

He pay cash = Rs 4000

Remaining balance = Rs 22000 — Rs 4000

= Rs 18000
Annual instalment = 1000 + Interest on unpaid amount @10%
unpaid amount instalment

1* Instalment 18000 = = x 18000 =1800 = 1000+1800= 2800
24 Instalment 18000 — 1000 =17000 = an x 17000 =1700 = 1000+1700 = 2700
3° Instalment 17000 — 1000 =16000 = a x 16000 =1600 = 1000+1600 = 2600

Misc 15 teachoo
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.
Number of letters

1% set 4

2” set 4x4=16

3° set 16x4=64

sth set
Hence, the sequence is
i.e. 4, 16, 64, 256, ..., upto 8th term

Misc 16 teachoo
A man deposited Rs 10000 in a bank at the rate of 5% simple interest
annually. Find the amount in 15" year since he deposited the amoun
and also calculate the total amount after 20 years
In simple interest, the interest remains same in all year.
Interest per year = 10000 x 5%
= 500

Hence,
Amountin 1% year = 10000
Amount in 2"? year = Amount in 1% year + Interest = 10000 + 500

= 10500
Amount in 3 year = Amount in 2™ year + Interest = 10500 + 500

= 11000

Misc 17 (Method 1) teachoo
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.
We use the formula
r n
a=P(1-<5)

Here, P = principal = 15625

r=rate of depreciation = 20%

n =number of years = 5 years

A be the depreciated value

Misc 18 teachoo
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.
Let total work = 1
and let total work be completed in =n days
Total K 1
Work done in 1 day = —__ 2 ore ors =
Number of days tocomplete work n
This is the work done by 150 workers
. 1
Work done by 1 worker in one day = ——
150n

teachoo.com
Misc 1
Show that the sum of (m + n)"" and (m—n)* terms of an A.P. is
equal to twice the m" term.
First we calculate (m+n), (m—n)* and m"™ terms of an A.P
We know that
a,=a+(n—1)d (2)
Where
a, isn‘ term of AP
a be the first term
& d be the common difference of the A.P.
For (m+n) term
puttingn=m+nin (1)
amin =at(mt+n-—1)}d ...(2)

teachoo.com
Misc 2
If the sum of three numbers in A.P., is 24 and their
product is 440, find the numbers.
Let the three numbers in A.P. be
a-d,a,atd.
Sum of the three numbers is 24
(a—d) + (a) +(a+d)=24
atat+tat+d-d=24
3a+0O =24
3a=24
a=8

teachoo.com
Misc 3
Let the sum of n, 2n, 3n terms of an A.P. be S,, S, and
S;, respectively . Show that S, = 3 (S,— S,)}
We know that
Sum of n terms = ; [2a + {n— 1)d]
ie. S, = ; [2a + (n - 1)d]
where ais the first term
& dis the common difference of the A.P.
[tis given that
Sum of nterms = S,
i.e. 5 [2a+(n—1)d] = S, AL}
Also,
Sum of 2n terms = S,
i.e. = [2a + (2n-1)d] =5, (2)

teachoo.com
Misc 4
Find the sum of all numbers between 200 and 400 which are
divisible by 7.
Numbers between 200 & 400 are
201, 202,203,... 398,399
Finding minimum number in 201, 202,203,... 398,399
which is divisible by 7
201 _ 598
7 7
202 _ 596
7 7
203 _ 99
7
So the series will start from 203

teachoo.com

Misc 5
Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
Sum of integers divisible by 2 or 5

= Sum of integers divisible by 2 + Sum of integers divisible by 5

— Sum of integers divisible by 2 &5

Finding sum of numbers from 1 to 100 divisible by 2
Integers divisible by 2 between 1 to 100 are

2, 4, 6, 8, ...100
This forms an A.P. as difference of consecutive terms is constant.
First term =a = 2
common difference d=4-2 =2
Last term = 1 = 100

teachoo.com

Misc 6
Find the sum of all two digit numbers which when divided
by 4, yields 1 as remainder.
Two digit numbers are
10,11,12,13,.......98,99
Finding minimum number in 10,11,12,13........98,99 which
when divided by 4 yields 1 as remainder

30 _ 42

4. 4

41 _ 43

4. °4

2,

4

13 _ 41

4.4
So the sequence will start from 13

teachoo.com
Misc 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.
Let the first four terms of A.P. be
a,atd,a+2d,a+3d
Given,
First term =a =11
Let common difference = d
Also given that,
Sum of first four term is 56
at (a+d)+{a+2d) + (a+ 3d) =56
11+ (11 +d) +(11 + 2d) +(11 + 3d) = 56
11+114+11+11+d+ 2d +3d = 56

toachoo.com
Misc 15
The p*, g'? and r" terms of an A.P. are a, b, c respectively.
Show that (q — r) a+(r — p)b+ (p-—q)c=0
Here we have small ‘a’ in the equation, so we use capital ‘A’ for
first term
We know that,
A,=A+({n-1)D
where A, is the n‘* term of A.P.
nis the number of terms
Ais the first term, D is the common difference
It is given that p'* term of an AP is a
ie.A,=a
Putting n =p
At(p-1)D=a
a=A+(p-1)D

toackoo.com
Misc 16
if a(> + *), b(= + *), (= + *) are in AP, prove that a,
b c c a a b
b, c are in AP
Given that a(> + ), b(= + ), (= + =) are in AP.
b c c a a b
i a(> + ), b(= + “), (= + *) are in AP
b c c a a b
Adding 1 to each term
a(> + =) +1, b(= + *)+1,e(2 + s)+1 are in AP
Bb c c a a Bb
a(> + *)+ £, b(= + *)+2, (= + s)+e are in AP
b c a c a b a db c
a(p +242), b(t +242), c(= + = + 2) are in ap
b c a c a b a Bb c

teachoo.com
Misc 20
: . 111
Ifa, b, care in A.P, ; b,c, d are in G.P and vere
are in A.P. prove that a, c, e are in G.P.
It is given that a, b, c are in AP
So, their common difference is same
b-a=c-b
b+b=cta
2b=c+a
batt .. (1)
2
Also given that b, c, d are in GP
So, their common ratio is same
clad
boc
c? = bd wa (2)

teachoo.com

Misc 23
Find the sum of the first n terms of the series:
34+74+13+21+31+..
It is not an AP or a GP
Let
§,=3+74+134+214+31+.4+a,,+a, (1)
$,=04+3+7+134+21+...+a, ,+a, +4, w= (2)
Subtracting (2) from (1}
S,—-S, = 3-0 + [(7 — 3) + (13 — 7) + (21-13) +... + (a, —-a,_>}

+(a,—a,_all—ay
0=3+[4+6+8+..a,,]-a,
a,=3+[4+6+8+.. +a,_,] (3)

teachoo.com
Misc 24
If S,, S,, S, are the sum of first n natural numbers, their
squares and their cubes, respectively, show that
95,2 = S, (1+ 8S,)
It is Given that
S, is the sum of n natural numbers
ie.$,=14+2+3+..4¢n
_n(n+t)
S, =
S, is the sum of square of n natural numbers
ie.S, =124+2?4+37..4+n?
S,= n(n+1)(2n+1)
27 6

teachoo.com
Misc, 25
Find the sum of the following series up to n terms:
Vo 1+2? 1942943°
1 143 14345
121342? 1°4+27+3°
1 44+3 ° 14345
n‘* term of series is
ac: +274 374.4 n?
nN 443454... +¢nterms
We solve denominator & numerator separately
134+ 234+ 334+ ..4+n°
n
n=1
_ (nespy’ (4)
= 5 ee

teachoo.com
Misc 26
1X 27+2x3?+..4¢nX(n+1)? 3n45
Show that DP 4 2kT wt nxt ye =.
Vx242°x3+..¢n°xX(n+ 1) 3n4+1
Taking L.H.S
1X 2?42x3?+..+¢nX (n+ 1)?
Vx242?x34+..¢n’x (+t 1)
We solve denominator & numerator separately
Solving numerator
Let numerator be
S,=1x2?4+2x3*4+..+n x(n+1p
n term is n x (n+ 1)?
= 2
Let a, =n(n+ 1)
=n(n?+1+2n)
=n? +n-+2n?

teachoo.com
Ex9.2,1
Find the sum of odd integers from 1 to 2001.
Integers from 1 to 2001 are 1, 2, 3, 4, ..........2001
Odd integers from 1 to 2001 are 1,3,5,...1999,2001
This sequence forms an A.P as difference between the
consecutive terms is constant.
So, the A.P. is 1,3,5,...1999,2001
Here
First term =a = 1
Common difference = d
=3-1
=2
& last term = 1 = 2001

teachoo.com

Ex 9.2,2
Find the sum of all natural numbers lying between 100 and 1000,
which are multiples of 5.
Multiples of 5 are 5, 10, 15,20,25 ......
Multiples of 5 between 100 and 1000 are
105, 110,115, ... ,990,995.
This sequence forms an A.P. as difference between the
consecutive terms is constant.
Here,
first term =a = 105
Common difference = d

= 110-105

=5
& last term = 1 = 995

teachoo.com
Ex9.2 ,3
In an A.P, the first term is 2 and the sum of the first five
terms is one-fourth of the next five terms. Show that 20"
term is —112.
It is given that
First term = a= 2
Also
Sum of first five terms = : (Sum of next 5 terms)
Sum of first five terms = : (Sum of 6 to 10" terms}
Sum of first five terms == ( Sum of first 10 terms )
4\ —Sum of first five terms.
1
Ss = 7(S10 - Ss)
4S, = Sig — Ss

teachoo.com
Ex 9.2,4
How many terms of the A.P. -6,— > —5... are needed
to give the sum —25?
. Ww

AP is of the form — 6, — ze Bae
Here
First term =a=—-6
Common difference = d = ->- (-6)

=-146

2

_ 11412

~ 32

=i

“2
& Sum of n terms = S, =—25
We need to find n

toackoo.com
Ex 9.2,5
In an A.P., if p* term ist and q'" term ist prove that the
sum of first pq terms is ; (pq +1) where p #q.
We know that
a,=at(n—1)d
Where a, is n™ term of AP,
nis the number of terms,
a be the first term
& d be the common difference of the A.P.
It is given that
* term is =
P q
lea, ==
ay =F
1
a+(p—1)d =5 (1)

teachoo.com

Ex 9.2,6
If the sum of a certain number of terms of the A.P. 25,
22, 19, ... is 116. Find the last term
AP is of the form 25, 22, 19, ..
Here
First term = a= 25
Common difference = d = 22-25 =-3
Sum of n terms = S, = 116.
We need to find last term a,
First, we find n
We know that

S, => [2a +(n—1)d]

Putting values

116 => [2 x 25 + (n-1)(-3)]

toackoo.com
Ex9.2 ,7 (Method 1)
Find the sum to n terms of the A.P., whose k' term is 5k + 1.
It is given that
k® term =5k +1
a, =5k+1,
Putting k = 1
a,=5x 141
=54+1
=6
Similarly,
Putting k = 2
ay=5x24+1
=10+1
=11

toachoo.com
Ex 9.2,8
If the sum of n terms of an A.P. is (pn + qn?}, where p and
q are constants, find the common difference.
Let a,, a), ... a, be the given A.P
Given, Sum of n terms = (pn+ qn}
S, =(pn + qn?) (1)

Putting n = 1in (1)
S,=(px 1+qx 1?)

=ptqxi1

=ptq
Sum of first 1 terms = First term
- First term = a, = S,

=ptq

teaehoo.com
Ex9.2,9
The sums of n terms of two arithmetic progressions are in the ratio
5n + 4: 9n + 6, Find the ratio of their 18" terms.
There are two AP with different first term and common difference
For the first AP
Let first term be =a
Common difference = d
Sum of n terms = S, = 5 [2a + {n— 1)d]
& n term =a, =a +(n—1)d
For the second AP
Let first term be =A
common difference = D
Sum of n terms = S, = > [2A + (n—1)D]
& n term =A, =A+(n-1)D

Ex 9.2, 10 ‘eatoocon
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.
We know that
S,=5(2a+(n-1)d)

Where, S, = sum of n terms of A.P.

n= number of terms

a = first term and d = common difference
Now,
Sum of first p terms = S,

= $ [2a +(p—1)d]

toachoo.com
Ex 9,2,11
Sum of first p,q,r terms of an A.P are a,b,c resp.
b
Prove that = (q-r)+ q (r- p)+ < (p-q)=0
Here we have small ‘a’ in the equation, so we use capital ‘A’ for
first term
We know that,
S,= 5 [2A +(n—1)D]
where S, is the sum of n terms of A.P.
nis the number of terms
A is the first term, D is the common difference
Given,
Sum of first p terms =a
S)=a
2 [2A +(p—1)D] =a
a=2[2A+(p—1)D]

toackoo.com
Ex 9.2 ,12
The ratio of the sums of m and n terms of an A.P. is m:n.
Show that the ratio of m“ and n'* term is (2m — 1): (2n— 1).
We know that
S,=5(2a+(n-1)d)
Where, S, = sum of n terms of A.P.
n= number of terms
a= first term and d = common difference
Thus, Sum of n terms = S,
= 3(2a + (n—1)d)
And
Sum of m terms = S,,
= (2a +(m—1)d)

tonchoo.com

Ex 9.2,13
If the sum of n terms of an A.P. is 3n? + Sn and its m™ term is
164, find the value of m.
Let a,, a,, ... a, be the given A.P
Given, Sum of n terms = 3n?2+5n

S, = 3n?+5n (1)
Putting n = 1in (1)

S,=3x1?4+5x 1

=3x14+5x1

=3+5

=8
Sum of first 1 terms = First term
~. First term =a, =S,

=8

toackoo.com
Ex9.2,14
Insert five numbers between 8 and 26 such that the
resulting sequence is an A.P.
We know that to insert n numbers between a & b
common difference (d) = 2—*
n+1
Here,
We need to insert 5 numbers between 8 and 26
So,b=26,a=8
& number of terms to be inserted =n =5
Therefore,
= 26=8
541
_18
~ 6
=3

toackoo.com
Ex9.2,15
a®+pr, .
If >>—azz is the A.M. between a and b, then find the
a + prt
value of n.
We know that arithmetic mean between a & b is
Am, =2t2
2
an al + pe
It is given that AM between a & b is Gig phi
So,
at+b™ _atb
ah-14pn-1 7 2
2(a" + b") = (a + b) (a"~ 1+ bn-4)
2a" + 2b" = a(a"~1+ b"~1) + b(at-14 b"-4)
2a" + 2b" = aa"-1 + ab"-14 ba"-1+ bbe?
2a" + 2b"=at.a®-t+ab.~1+ba"-1+bt.b.-!
2a" + 2b" = attn- 14 gb"-14 ban-14 p1+n-1

teachoo.tom
Ex 9.2,16
Between 1 and 31, m numbers have been inserted in such a
way that the resulting sequence is an A.P. and the ratio of
7‘ and (m— 1)" numbers is 5 : 9. Find the value of m.
We know that to insert n numbers between a & b
: b-
common difference (d) = “—<
nti
Here,
We have to insert m numbers between 1 and 31
S0,b=31,a=1
& number of terms to be inserted =n=m
Therefore,
dette}
m+1
_ 30
~ m+

toackoo.com
Ex9.2,17
Aman starts repaying a loan as first instalment of Rs.100.
If he increases the instalment by Rs 5 every month, what
amount he will pay in the 30" instalment?
First instalment = 100.
Every month instalment increases by Rs 5
-. Second instalment = 100 +5
=105

Third instalment = 105 +5

=110
So, the instalments are 100, 105, 110, ...
The instalments is in A.P as difference between
consecutive terms is constant.

teachoo.com
Ex 9.2,18
The difference between any two consecutive interior angles of a
polygon is 5°. If the smallest angle is 120°, find the number of
the sides of the polygon.
Difference between consecutive angles = 5°
Smallest angle = 120°
Second smallest angle = 120° + 5°
=125°
Third smallest angle = 125° + 5°
= 130°
Thus, the angles are 120, 125,130.....
The angles form an A.P. as difference of consecutive terms is
constant.
Here,
first term = a= 120
Common difference = 125-120 =5

teackoo.com
Ex 9.4, 1
Find the sum to n terms of the series
1x24+2x34+3x4+4x5+.,.

1% term 1x2

2°¢ term 2x3

3 term 3x4

n term nx (n+1)
Step 1: Find n'" term (a,)
Here, a, =n(n+1)

=mrtn

Step 2: Finding sum of n terms

teachoo.com

Ex 9.4, 2
Find the sum to n terms of the series
1x2x342x3x4+3x4x54,..
1x2x34+2x3x44+3x4x5t,..
Step 1: Find n' term (a,)
Here, a, =n (n+1)(n+ 2)

=(n? +n} (n+ 2)

=n?(n + 2)+n(n + 2)

=n+2n?+n?42n

= n+ 3n? + 2n
Step 2: Finding sum of n terms

n
S.=). an
n=1

teachoo.com
Ex 9.4, 3
Find the sum to n terms of the series
3x 1245x2747 324+...
1t term 3}x|1
2°¢ term 5)xl2
3 term 7/3?
n™ term (nth|term of 3, 5, 7, ...) x (n™ term of 12,22,32,..)
n‘ term of 3, 5, 7, +. nt" term of 12, 22, 32, ...
This is AP
where First term = a = 3 nth term = n2
Common difference = d =5-—3=2
n“ term =a + (n—-1)d
=3+(n—1)2
=3+2n-2
=2n+1

teachoo.com
Ex 9.4, 4
Find the sum to n terms of the series
1 1 1
1x2 2x3 3x4
+ 1 1 1
n term of series —~ + —~4+—— ta
1X2 2x3 3x4
is +
n(n +1)
>a,= 1
0 n(n +1)
_M@+ij-n
~ n(n +1)
_ @+1) n
~ n(w+1) n(nt1)
1 a
= -— (1)
n ™+tti1)

teachoo.com
Ex 9.4, 5
Find the sum to n terms of the series
524 62+ 72 + eee + 202
524 62+ 72 + eee + 202
= (12+ 224 3% 4+ 47+57+ 62+... + 202) — (12+ 2? + 3? + 47) (4)
We know that
Sum of square of n natural number is
ie. (127427 4.0.4 2) = RetNEnst) (2)
6

For 12+ 22+... + 207
n=20
Putting n = 20 in (2)
4224. £202 = LOGO+ ECO) +1)

_ 20(21)(40 + 1)

~ 6

teachoo.com
Ex 9.4, 6
Find the sum to n terms of the series
3x8+6x11+9x 14+...
1 term 3}x|8
2°¢ term 6)x!11
3 term 9)x/14
n™ term (nth |term of 3, 6, 9...) x (n' term of 8, 11, 14, ...)
n term of 3, 6, 9, ... n‘* term of 8, 11, 14, ...
This is AP This is AP
where First term = a = 3 where First term = a = 8
Common difference = d = 6 — 3 = 3] Common difference = d = 11-8 =3
n* term =a +(n—1)d n“ term of AP=a+(n—1)d
=3+(n—1)3 =8+(n-1)3
=34+3n-3 =8+3n-3
=3n =3n+5

teackoo.com
Ex 9.4, 7
Find the sum to n terms of the series
124 (124+ 22) + (12427432) 4..
Step 1: Find n' term (a,)
1t term 1?
2°4 term (12 + 2?)
3 term (12+ 22437)
n term (12 + 2? 43? 4.....4n?)
Here a, = (17+ 27+ 3?4.......4.n?)
n(n+1)(2n+1)
a, =
6
_ @? +n)(2n +1)
~ 6
_ (20? +n? + 2n? +n)
~ 6
_ Qn? + 3n? +n)
~ 6

teackoo.com
Ex 9.4, 8
Find the sum to n terms of the series whose n' term is given
by n(n +1) (n+ 4).
Given
a, =n(n +1) (n+ 4)
=(n?+n)(n + 4)
=n? (n+4)+n(n+4)
=n? +4n?+n?+4n
=n?4+n?+4n’*+4n
=n? +5n?+4n
The sum of n terms is
n
5,.=). an
n=1
n
= ». n3+ 5n244n
n=l

teackoo.com
Ex 9.4, 9
Find the sum to n terms of the series whose n" terms is
given by n2 + 2"
Given
a, =n? +2"
Now, sum of n terms is
n
Sy -) an
n=1
n
n=1
nr n
n=1 n=l
n
We will solve > 2 separately
n=1

teachoo.com
Ex 9.4, 10
Find the sum to n terms of the series whose n™ terms is
given by (2n— 1)?
Given
a, = (2n— 1)?
=(2n) + (1)? — 2(2n)(1)
=4n?+1-4n
=4n?-4n+1
Sum of n terms is
n
s.=). an
n=1
n
= » 4n?-4n+1
n=1
n n n
= D4 Sant D1
n=1 n=l n=1