SEQUENCE
A collection of numbers arranged in a definite order according to some definite rule (rules) is called a sequence.
Each number of the sequence is called a term of the sequence. The sequence is called finite or infinite according as the number of terms in it is finite or infinite.
Meaning of Sequence
A sequence is a group of objects which follow some particular pattern. If we have some objects listed in some order so that it has 1 term, 2 term and so on, then it is a sequence.
Example
In this image we can see the group of sticks with a certain pattern, so we can observe that what could come next
As we can see that the next figure in the sequence will have 9 sticks.
ARITHMETIC PROGRESSION
A sequence is called an arithmetic progression (abbreviated A.P.) if and only if the difference of any term from its preceding term is constant. It could be in ascending or descending form according to the constant number.
A sequence in which the common difference between successors and predecessors will be constant. i.e. a, a+d,a+2d
This constant is usually denoted by ‘d’ and is called common difference.
Example
Here we are getting the terms by adding 3 every
time. this is the difference between the two
successive terms so it is called the difference.
The difference is represented by “d”.
In the above example we can see that a1 =0 and a2 = 3.
The difference between the two successive terms is
a2 – a1 = 3
a3 – a2 = 3
If the first term of an arithmetic sequence is a and the common difference is d, then the nth term of the sequence is given by:
tn = a + (n−1) d
NOTE : The common difference ‘d’ can be positive, negative or zero.
SOME MORE EXAMPLES OF A PARE
(a) The heights (in cm) of some students of a school standing in a queue in the morning assembly are 147, 148, 149, ….. ,157.
(b) The minimum temperatures (in degree celsius) recorded for a week in the month of January in a city, arranged in ascending order are 3. 1, — 3. 0, — 2. 9, — 2. 8, — 2.7, — 2. 6,— 2. 5
(c) The balance money (in ) after paying 5% of the total loan of Z 1000 every month is 950, 900, 850, 800, ….50.
(d) The cash prizes (in ₹) given by a school to the toppers of Classes Ito XII are, respectively, 200, 250, 300, 350„ 750.
(e) The total savings (in ₹) after every month for 10 months when Z 50 are saved each month are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500.
nᵗʰ TERM OF AN A.P. : It is denoted by t n and is given by the formula, tₙ = a + (n —1)d
where ‘a’ is first term of the series, n is the number of terms of the series and ‘d’ is the common difference of the series.
NOTE : An A.P which consists only finite number of terms is called a finite A.P. and which contains infinite number of terms is called infinite A.P.
REMARK : Each finite A.P has a last term and infinite A.Ps do not have a last term.
RESULT: In general, for an A.P a₁ , a₂ , ..., aₙ , we have d= tk+1 — tk where tk+1 and tk are the (k+1)ᵗʰ and the kᵗʰ terms respectively.
SUM OF FIRST n TERMS OF AN A.P.
It is represented by symbol Sn and is given by the formula, Sn = n 2 n 2
REMARK : The nth term of an A.P is the difference of the sum to first n terms and the sum to first (n — 1) terms of it.
i.e. an = Sn — Sn-1
TO FIND nth TERM FROM END OF AN A.P. :
nᵗʰ term from end is given by formula l – (n – 1)d
nth term from end of an A.P. = nth term of (l, l — d, l – 2d, …….)
=l+(n-1)(—d)=l—(n-1)d.
PROPERTY OF AN A.P. :
If ‘a’ , b, c are in A.P., then
b — a= c — b or 2b= a + c
THREE TERMS IN A.P. :
Three terms of an A. P. if their sum and product is given, then consider
a—d,a,a+d.
FOUR TERMS IN A.P. :
Consider a —3d, a — d, a+ d, a +3d.
NOTE :
The sum of first n positive integers is given by Sn = n(n + 1) / 2
Geometric Sequence
Any sequence in which the ratio between every
successive terms is constant then it is called
Geometric Sequence. It could be in ascending or
descending form according to the constant ratio.
Example
Here
t1 =1
t2 = 4 = t1 × 4
t3 = 16 = t2 × 4
Here we are multiplying it with 4 every time to get
the next term. Here the common ratio is 4 .
The ratio is denoted by “r”.
Geometric Progression(GP) or Geometric
Sequence is sequence of non-zero numbers in
which the ratio of any term and its preceding
term is always constant.
A geometric progression(GP) is given by a,
ar, ar², ar³, ...
where a = the first term , r = the common
ratio
Examples
1, 3, 9, 27, ... is a geometric progression(GP)
with a = 1 and r = 3
2, 4, 8, 16, ... is a geometric progression(GP)
with a = 2 and r = 2
If a, b, c are in GP, b² = ac
nth term of a geometric progression(GP):
tn = arn-1
where tn = nth term, a= the first term , r =
common ratio, n = number of terms
Sum of first n terms in a geometric progression(GP):
sn = where a= the first term,
r = common ratio,
n = number of terms
Geometric Mean
If three non-zero numbers a, b, c are in GP, b is
the Geometric Mean(GM) between a and c. In
this case, The Geometric Mean(GM) between two
numbers a and b = sqrt(ab)
(Note that if a and b are of opposite sign, their GM is not defined.)
Additional Notes on GP
To solve most of the problems related to GP, the
terms of the GP can be conveniently taken as
3 terms:
5 terms:
If a, b, c are in GP, In a GP, product of terms equidistant from beginning and end will be constant.
Power Series :
Important formulas
1 + 1 + 1 + ⋯ n terms = ∑ 1 = n
Problems
1. Define sequence and series?
Ans: A list of numbers in definite order is called sequence.
eg. 3,5,7..............
When the terms of sequence are connected by addition and subtraction signs e.g. 2+4+6+8..............., it is a series.
2. If 4, a and 16 are in the geometric sequence, find the value of a.
Solution:
Here,
4, a 16 are in GS
So, GM = sqrt (AB)
⇒ a = sqrt (4 × 16)
3. If the arithmetic mean between 4 and x is 34, find their geometric mean.
Solution:
Here,
AM =34
We know that, AM =
⇒ 34 =
⇒ 64 = x.
Now,
GM =
= √(x
=
= ±16
Thus, the GM is 16.
4. If the geometric mean 2 and x is 4, find their arithmetic mean.
Solution:
Here,
a =2, b =2 and GM = 4.
So,
⇒ 2x = 16
∴ x = 8
Now, AM =
Thus, the AM is 5.
5. If m+2 , m+8 and 17+m are in geometric sequence, find the value of m.
Solution:
Here,
the team of a GS, are: m+2 , m+8 and 17+m
So,
(m+8)2 = (m+2) (17+m)
⇒ m²+16m+64=17m+m²+34+2m
⇒ 16m +64=19m+34
⇒ 30 =3m
∴ m = 10.
Thus, the value of m is 10.
6. Which term of the series 3+6+12+.........is 192 ? Find it.
Solution:
In the given series
first term (a) = 3 and
common ratio(r) = 12/6 = 2
Now, tn = arn-1⇒ 192 = 3× 2n-1
⇒ 64=2n-1
⇒ 26 = 2n-1
⇒ n-1= 6
∴ n = 7.
Thus, 7th term of series is 192.
7. Find the number of terms of the given series. 2-2√
Solution
Here, 2-2√2 +4 - 4 √2 +........+16 is a geometric series.
first term (a) =2
Common ratio (r) = ⇒ arⁿ`¹ = 16
⇒ 2⨉(-√2) n-1 = 16
or, (-√2) n-1 = 8
⇒ (-√2) n-1 = (-√2) 6
⇒ n-1=6
∴ n = 7.
Thus, there are 7 terms.
8. Which term of the series 5+9+13+.........is 85? Find it.
Solution:
Here,
5+9+13+.........and tn = 85
We know that, a = 5
d= t2 - t1 = 9-5=4.
By the formula, tn = a+(n-1)d
⇒ 85= 5+ (n-1)4
⇒ 80= (n-1)4
⇒ 20= n-1
∴ n = 21
Thus, 21st terms is 85.
9. If the third term of a geometric series is 3. Find the product of its first five terms.
Solution:
Here,
Third term (t3) = 3
Now, production of first five terms
= a × ar × ar2 × ar3 × ar4
= a5r10
= (ar2)5
= (3) 5
=243.
Thus, the product of first five term is 243.
10. Find the arithmetic mean and geometric mean of 3 and 27.
Solution:
Here,
One term (a) = 3 and other term (b) =27.
We have, Arithmetic mean (AM) =
=
= 15.
Geometric mean (GM). =
= Sqrt(3×27)
= ± 9
Thus, the AM and GM are 15 and ± 9 respectively.
11. There are five geometric means between a and b. If second and last mean are 2 and 16 respectively, find the values of a and b.
Solution:
Here,
a ,m₁ ,m₂ (= 2), m₃ ,m₄ ,m₅ (=16) ,b
We have,m2 = 2
i.e. ar2 = 2 ……… (i)
Again, m5 = 16
i.e. ar5 = 16 ……… (ii)
Dividing (ii) by (i) then,
⇒ r3 = 8
⇒ r3 = 2³
∴ r = 2
Substituting r=2 in eqⁿ 1, we get
a × 22 = 2
i.e. a =
Now, b = t7 = ar6 =
∴ b = 32
Thus, the values of a and b are
12. If the sum of the first 9 terms of an arithmetic series is 72 and the sum of the first 17 terms is 289 then the sum of the first 25 terms.
Solution:
Given,
S9 = 72
⇒
⇒ a + 4d = 8
⇒ a = 8 - 4d ……….. (i),
and S17 = 289
i.e.
⇒ a + 8d = 1
⇒ 8 - 4d + 8d = 1
⇒ d =
Substituting the values of d in (i), we get
a = -1
Now,
Sn =
∴ S25 =
=
=
= 650.
Thus, the sum of 25th term is 650.
14. If the second term and fifth term of a geometric series are 15 and 405 respectively, find the sum of the series from first to 6 terms.
Solution:
Given,
Second term (t2) = 15
i.e. ar = 15
and fifth term (t5) = 405 .
i.e. ar4 = 405
⇒ar.r³ = 405
⇒15 r³ = 405
⇒r³ = 27
⇒r = 3
And hence a = 5
Now, sn =
∴ s₆ =
=
=
= 1820.
Thus, the sum of the series from first to 6 terms is 1820.
15. Find the sum of the first 9 terms of a geometric series whose third term and seventh term are 20 and 320 respectively.
Solution:
Here,
Third term (t3) = 20
i.e. ar2 = 20
& Seventh term (t7) = 320
i.e. ar6 = 320
⇒ ar².r⁴ = 320
⇒ 20 . r⁴ = 320
⇒ r⁴ = 16
⇒ r = ±2
And hence a = 5.
Now, sn =
Using r = -2, sum of first 9 term is:
s₉ =
=
=
= 5 × 171
= 855
Again
Using r = 2, sum of first 9 term is:
s₉ =
=
=
= 2555
Thus, the sum of first 9 terms is 855 or 2555.
5 and 25.
Solution:
Here,
5....m1….m2…m3….m4……25
First term (a) = 5
Sixth term (t6) = 25
i.e. a + 5d = 25
⇒ 5 + 5d = 25
∴ d = 4.
Now,
m1 = a + d = 5 + 4 = 9
m2 = a + 2d = 5 + 2 × 4 = 13
m3 = a + 3d = 5 + 3 × 4 = 17
m4 = a + 4d = 5 + 4 × 4 = 21.
Thus, the four means are 9, 13, 17 and 21.
17. There are n arithmetic means between 4 and 24. If the ratio of third mean to that last mean is 4:5 then find the number of terms in the series.
Solution:
Here,
First terms (a) = 4,
Last term (b) = 24.
We know that,
d =
We have that,
⇒
⇒
⇒
⇒
⇒
⇒ 24n+ 4 = 5n + 80
⇒ 19n = 76
∴ n = 4.
Thus, the total numbers of terms = 4 + 2 = 6.
18. The 3rd and 7th terms of an arithmetic series are 18 and 30 respectively, find the sum of the first 20 terms.
Solution:
Here,
Third term (t3) = 18
i.e. a + 2d = 18 (∵ tn = a + (n – 1)d )
⇒ a = 18 - 2d ..............(i)
Seventh term (t7) = 30
i.e. a + 6d = 30
⇒18 - 2d + 6d = 30
⇒ d = 3
Substituting it in eqⁿ (i), we get:
a = 18 - 2×3 = 12
Now,
Sn =
∴ S20 =
=
= 10 × 81
= 810.
Thus, the sum of first 20th term is 810.
19. Find the sum of the first 8 terms of a geometrical series whose fourth term is 54 and common ratio is 3.
Solution:
Here,
Common ratio (r) = 3 ,
Fourth term (t4) = 54
i.e. ar3 = 54 (∵ tn = arn – 1 )
⇒a . 33 = 54
∴ a = 4.
Now
Sum of terms (Sn) =
∴ Sum of 8 terms (S8) =
=
= 38 – 1
= 6561 – 1
= 6560
Thus, the sum of 8 terms is 6560.
20. Three terms in an arithmetic progression have sum 21and product 315. Find the terms.
Solution:
let the three terms of arithmatic progression are a – d, a, a + d respectively.
Given,
a – d + a + a + d = 21
⇒ 3a = 21
∴ a = 7 .
and
Product = 315
⇒ (a – d)(a + d). a = 315
⇒ (7 – d) (7 + d) × 7 = 315
⇒ (49 – d2) = 45
⇒ 4 = d2
∴ d = ± 2 .
When a = 7 and d = -2 then,
Three terms, (a – d), a, (a + d) = (7 + 2), 7, (7 – 2) = 9, 7, 5 .
When a = 7 and d = 2 then,
Three terms = (a – d), a, (a + d) = (7 - 2), 7, (7 + 2) = 5, 7, 9.
Thus, the three terms are 5, 7, 9 or 9, 7, 5.
21. The ninth and nineteenth terms of an Arithmetic Series are 40 and 60 respectively. Find the sum of the first 25 terms of the series.
Solution:
Here,
Ninth term (t9) = 40
i.e. a + 8d = 40
⇒a = 40 - 8d ......... (i)
And Nineteenth term (t19) = 60
i.e. a + 18d = 60
⇒ 40 - 8d + 18d = 60
⇒ 10d = 20
∴ d=2
substituting it in equation (i), we get:
a = 40 - 8*2 = 24
Now,
Sn =
∴ S25 =
=
=
Thus, the sum of 25th term is 1200.