SEQUENCE AND SERIES

SEQUENCE
Is a set of numbers written in a definite order such that there is a rule by which the terms are obtained.

Or
Is a set of number with a simple pattern.
Example
1. A set of even numbers
• 2, 4, 6, 8, 10 ……
2. A set of odd numbers
• 1, 3, 5, 7, 9, 11….
Knowing the pattern the next number from the previous can be obtained.
Example
1. Find the next term from the sequence
• 2, 7, 12, 17, 22, 27, 32
The next term is 37.
2. Given the sequence
• 2, 4, 6, 8, 10, 12………
What is
i)The first term =2
ii) The 3rd term =6
iii)The 5th term =10
iv)The nth term [ the general formula]
2=2×1
4=2×2
6=2×3
8=2×4
10=2×5
12=2×6
Nth =2xn
Therefore nth term =2n
Find the 100th term, general formula =2n
100th term means n=100
100th term =2×100
100th term =200
3. Find the nth term
ScwvzDb0QsJALyAfsRCUBXKS WWcji4d7hisftGe YGqyf8Iu TZ LzT4MMzKcV LF4Sg0l94eIKGoOuj8b YVLQPSjUNbaBCOh9Dd7qqQcHqhGnr3LII NEnZD9PTbnx1DS3yU ,Cv7jyLLrwZx0OCZFN8BaOePxG8I83OCn8P07XdH5lNzpAsFzmj7t2lvaOwUEWdYp4T9XHkFlubKpDilk4LvgO7sSHPvfeqlCbXDODupcOWCIdDRfD0 C7vzEzlmXVAT4LXnguYU

Solution
ScwvzDb0QsJALyAfsRCUBXKS WWcji4d7hisftGe YGqyf8Iu TZ LzT4MMzKcV LF4Sg0l94eIKGoOuj8b YVLQPSjUNbaBCOh9Dd7qqQcHqhGnr3LII NEnZD9PTbnx1DS3yU ,RsHInCuQbBFJ3fMM8DT7FxZ3 Dl9Wdocque0hxMJoZjRtpb4ShOUQnjzpCMEZyyA0VKrNo84rtRIUTyepoy0Vx373Sfbvc TQ8uH7kCVzg 2U1zhBf2BsJAeOVTQmkihMCQUe 4
4. Given the general term 3[2n]
a) Find the first 5terms
b) Find the sum of the first 3 terms
Solution
3[2n] when;
n=1, 3 [21] =6
n=2, 3 [22] =12
n=3, 3 [23] =24
n=4, 3 [24] =48
n=5, 3 [25] =96
First 5 terms = 6, 12, 24, 45, 96
Sum of the three terms
Sum of the first three = 6+12+24
Sum of the first three = 42

Exercise 1.
1. Find the nth term of the sequence 1, 3, 5, 7…..
2. Find the nth term of the sequence 3, 6, 9, 12……
3. The nth term of a sequence is given by 2n+1 write down the 10th term.
4. The nth term of a certain sequence is 2n-1 find the sum of the first three terms.
5. If the general term of a certain sequence is B4 Fnq KhtSfrYhp3o1y8jxhzTxdjvt0CyCPoTcs1EItYn64uszuIUn4yRmXDeiRaAcV TfHKv6jP683JeaOGD9B81K2ubjnQWvzS4B AsSMnQwp5MziY4PuoBp3H07PY6tjg2A
Find the first four terms increasing or
decreasing in magnitudes




Solution
1. 1, 3, 5, 7…nth
From the sequence the difference between the consecutive terms is 2 thus
nth =2+n


2. 3, 6, 9, 12……..nth
The difference between the consecutive terms is 3 thus
nth =3n
3. 10th = 2[10+1]
10th =20+1
10th =21
4. When
n=1, 21-1
n=2, 22-1
n=3, 23-1
Sum of the first three terms =1, 2, 4
Sum = 1+2+4
Sum = 7


5. When
n = 1, G2TIkvUG 5tMsM4cD GTrmMWqy9sd2xkVwANyn L7N 5aris4166 Bsy ZY BgFU5DVUXowpz PryaBaMOf6ZSMpQSlxLSpuQZQevvOSf2 U4KhXUNazMYwSebC79HPbe8HjjZc = 1
n = 2, 9F0aHC0qEiV6SA4i40A1yr6pDZe8Fd8vVQHP366cV1Rlem7ekhUWx80JBOgxWs OscyxPacD1p6M UPV4PZmeXyv8bdDChlnBfmsW9H8Q3uGkDuowyzZ1Oh2bn8sUWKbSY4cAZw =Hsn ASdGhIe6xlrzUET7uUHTS24NzSmGlr4EFtbZoNTuoAxk 8Y UvRUqiMO7sPfE7116QKyOozUdI5FWYWnTFxgOtfHSC3Z2sMsG4BykkgcubdHkbpiPskB9EH3sSiYWyu9Szg
n = 3, D8hYOoDzr7pUbkpOm33TVSoCA5EcZLTdwaiW4oGIp1YzFAPMRD1yuRIzAYZBzieUlRt7fpDZU7addUrxIWbCzl53O9zGHdNh1l4KzM58gxRfJKJwKfyl1PA1slGVgvdAQATvWPw = 1
n = 4, 3RuqUr8DZEf8sStUMmtMqeBbxexILDgx94tY4UN7vhsI1zaRG6uJG99 PkpgEd6FDDITRFCIsx9 UAZAZIeqd Ch9UPo8vHNSuwJAgdF7DvnodU1er6N93NYuBfc0rrBOWCBxQo = Hsn ASdGhIe6xlrzUET7uUHTS24NzSmGlr4EFtbZoNTuoAxk 8Y UvRUqiMO7sPfE7116QKyOozUdI5FWYWnTFxgOtfHSC3Z2sMsG4BykkgcubdHkbpiPskB9EH3sSiYWyu9Szg
The first four terms are UGMQbTxiS0CI IwW6S U0wMCmrRT0Zu7i89RMwDrwolcHIqEqhO4pGqwuKJPIFb Nr3yOfnpZJkMeefVYxZmRMTQ I7lPhbCO2 FefuK DDkJCGeYG R SLu OiRIAaGINbwU0
SERIES

Defination: When the terms of a sequence are considered as the sum, the expression obtained is called a serier or a propagation
Example
(a). 1+2+3+4+5+……….
(b). 2+4+6+8+10+………+100.
(c). -3-6-9-…….


The above expression represent a series. There are two types of series

1.Finite series
Finite series is the series which ends after a finite number of terms
e.g. 2+4+6+8+………….+100.
-3-6-9-12-………….-27.
2. infinite Series
Is a series which does not have an end.
e.g. 1+2+3+4+5+6+7+8…………..
1-1+1-1+1-1+1-1+1…………..

Exercise 5.2.1
1. Find the series of a certain sequence having 2(-1)n as the general term
2. Find the sum of the first ten terms of the series -4-1+2+…….
3. The first term of a certain series is k. The second term is 2k and the third is 3k. Find
(a). The nth term
(b). The sum of the first five terms

Exercise 5.2.1 Solution

1. n=1, 2(-1)1 = -2
n=2, 2(-1)2 = 2
n=3, 2(-1)3 = -2
n=4, 2(-1)4 = 2
n=5, 2(-1)5 = -2
The series is -2+2-2+2-2+2-2+2-2+2+………………..+2(-1)n

2. The sum of the first n terms of the series -4-1+2+5+8+11+14+17+20+23
= 95
3. (a) k + 2k + 3k + 4k +………. + nk
The nth term of the series is nk
(b) k + 2k + 3k + 4k + 5k = 15k
... The sum of the first 5 terms = 15k


ARITHMETIC PROGRESSION [A.P]

An arithmetic progression is a series in which each term differ from the preceding by a constant quantity known as the common difference which is denoted by “d“.
For instance 3, 6, 9, 12….. is an arithmetic progression with common difference 3.


The nth term of an arithmetic progression
If n is the number of terms of an arithmetic progression, then the nth term is denoted by An
Therefore An+1= An + d
e.g. First term = A1
Second term = A1+ d = A2
Third term = A2 + d = A3

Consider a series 3+6+9+12+…..
A1 = 3, d= 3
A2 = A1 + d
A3 = A2 + d = A1+d+d = A1+ 2d
A4 = A3 + d = A1+ 2d+ d = A1 + 3d
A5 = A4 + d = A1+ 3d+ d = A1 + 4d
A6 = A5 + d = A1+ 4d+ d = A1 + 5d

An = A[n-1] + d = A1+ (n-1)d

... The general formula for obtaining the nth term of the series is
An = A1+ (n-1)d

The general formula for obtaining the nth term in the sequence is also given by An =A1+[n-1]d

Question
1. A7 = C5XexQEcqcbT4vOMOExrCHk RXAb7kjKa2XR6Y9CztEOVRdEpfKW4HPiYK8D EbGx7w3qlyEPx6YXObiJ5NGQGRj52lzbr7nyVn0UtcQifWg YEA1brDUWj1122Yzj8Hr0zxjFsA2 and A4 =16. Find A1 and d.
Solution

A7= A1+6d
=A1+6d= JtTLFUw1KpbqSnDQy HU0gIXhT XbG0tLSKbnIqH8HoFM6tPJavlHyre HbMBVPVly2EwTkbKIQyTtu3bWZYdkbZder0GVSEzfUDLkrdxWF KKLPXBXUuiPV5Yfp6fra00C5aHo ( A1+d)
2A1+12d=5A1+5d
3A1=7d
A1=VsjKrt05QX0itvGY0xFVoxfy2REWwSONnVdExt APQ4UZ5zm0 TWV WQAOsbvzQCEJMVSSfaLJ BWUDtwDozDJpJlGdJDY6zxgNteW2TAYHURWfoMBK5 W5tp0 GWi6vaDEm474 d
A 4 =16
A4=A1+3d = 16
LM37O Efc29Mm2tY537dc QyN5Jn 47xSlkY57cIec2WvMcWJsA8HeI7dXDkx3e6ZsJMoPrJzlMI9XiO4 FvW0NPVoiWdTubLeDjPru8wmWhdqpxbU90pULViXUI49o5BxEBWnI d +3d= 16
CE3Sv MKhC8zs3uvJV3Gkabh6C2PV29t7vdEeH8b4Ve XuXbfQ7078GpGnQeBZV9do9oRyi IOAL8c Fs0Ulvfo9x 0FNsmCgRQOq6Bd9sFeEiiqDkWUJtXRWLZuuyBMQG5pyOU d = 16
d= 3
But A1 +3d= 16
A1 +9=16
A1= 7
Exercise 2
1. The pth term of an A.P is x and the qth term of this is y, find the rth term of the same A.P
2. The fifth term of an A.P is 17 and the third term is 11. Find the 13th term of this A.P.
3. The second term of an A.P is 2 and the 16th term is -4 find the first term.
4. The sixth term of an A.P is 14 and the 9th of the same A.P is 20 find 10th term.
5. The second term of an A.P is 3 times the 6th term. If the common difference is -4 find the 1st term and the nth term
6. The third term of an A.P is 0 and the common difference is -2 find;
(a) The first term
(b) The general term
7. Find the 54th term of an A.P 100, 97, 94
8. If 4, x, y and 20 are in A.P find x and y
9. Find the 40th term of an A.P 4, 7, 10……..
10.What is the nth term of an A.P 4, 9, 14
11.The 5th term of an A.P is 40 and the seventh term of the same A.P. is 20 find the
a) The common difference
b) The nth term
12. The 2nd term of an A.P is 7 and the 7th term is 10 find the first term and the common difference
Exercise 2 Solution
1. d = y-x
rth = A1+[n-1]d
= A1+nd-d
= A1+n[y-x]-[y-x]
= A1+[ny-nx-[y-x]
=A1+[ny-y]-[nx-x]
A1+g[n-1]-x[n-1]
rth=A1+[y-x][n-1]

2.
A5=17
A3=11
A13=?
A5 =A1+4d = 17
A3=A1+2d =17
A1+2d=11
Solve the simultaneous equation by using equilibrium method.
A1+4d=17
A1+2d=11
0H2UDvYB5FTPFVtg2Bg61uvr2c62jd0tNsZ0ZXEvyhvcrfinC8 V6p IM6vUucmS1Zl2 2 9uPbOXA3DGyMjOoGHP03V9nF60BC1biVodHaJBIuvx16yZwJXBo 4MCia7vGVNvY= HknBOlNb9blh8CR5MlK J0VN2G1iEcwmCszpFBltW7tcId35ZANN9 BqSSU2Em8L7ZCPCkVk7p0ZEaf1unj7PKZG6Tyn39fCmX4yExtt1GMEwpprz7LjenTQ9FO7tTjD9s3 0Y
d = 3
A1+4[3]=17
A1+12=17-12
A1=5
A13=A1+12d
A13=5+12[3]
A13=5+36
A13 =41
3.
A2=2
A16=-4
A1=?
A2=A1+d
A1+d=2
A16=A1+15d
A1 +15d= -4

Solve the two simultaneously equations by using the elimination method
A1 + d=2
A1 +15d=-4
4HOPkGBZsNwjLaenBMucckVC0EGOSlTtyfPds JKMAAAVt1Byb3xlemd23gMnOxE9KTLcQoa1GPDywE4EBcloqyr4Yt93 MJvB8TOnFy1VcplaDg2UhTgnjZd8db0WMcrFpc 34
d= TedsBCVvPPV2ksuxGtooN8VkGK43ni9b0y4CVpmNEAe WGXvU9F XwDn36rxN7ggTZWe3pkr R9M ApraW2HM13VBcDdfXDTppXdziCmPTT97i7byeDocLltIyYLJKlMCX6D 4c KYGO9ndr0y1M0JAs31cERmtQHNC4Wa RODVVLe440yRuBKREJc8XJ70uWml GH FFwG9KaHV8psRNoAFoWaMunsf94GQlVSrJxFlTVZq7PwQ541ckUMJ69CxxgfJcvuec GwiAI
From the 1st equation
A1 +TedsBCVvPPV2ksuxGtooN8VkGK43ni9b0y4CVpmNEAe WGXvU9F XwDn36rxN7ggTZWe3pkr R9M ApraW2HM13VBcDdfXDTppXdziCmPTT97i7byeDocLltIyYLJKlMCX6D 4c KYGO9ndr0y1M0JAs31cERmtQHNC4Wa RODVVLe440yRuBKREJc8XJ70uWml GH FFwG9KaHV8psRNoAFoWaMunsf94GQlVSrJxFlTVZq7PwQ541ckUMJ69CxxgfJcvuec GwiAI= 2
A1 = 2+ KYGO9ndr0y1M0JAs31cERmtQHNC4Wa RODVVLe440yRuBKREJc8XJ70uWml GH FFwG9KaHV8psRNoAFoWaMunsf94GQlVSrJxFlTVZq7PwQ541ckUMJ69CxxgfJcvuec GwiAI
A1 = GV80va6JuNe3StkWmGjPAQWQ0sZtTgdlsB1kBkYN 4MCsuJxTtkPG8YkH3 HY4urRowo3K9yZ2f71HrdyL6uEVix TLQ8bl1xE2kKnMxYJwSUAVoAMAZDpMM3B5Q1OQiWjm0UIc


4.
A6= 14
A9=20
A6= A1+5d = 14
A9 = A1+8d=20
Solve the simultaneous equation by using the elimination method
A1+5d= 14
A1+8d=20 ( difference between two equations)
 2idXQCA2JooWltEe Zyrd5OQ0AWD4MXVPg7AKW94ZxLdOQtw6e92sefhoEnjDdzCnTJuAofWdCrqKqWNiDpJqw6Uw9 FVTnxnuVxSpA2HMIwn DjfFxy4cCEmo1G JqnXK N6c
Then d = 2
From 1st equation
A1 +5[2] = 14
A1 +10= 14-10
A1=4
A10= A1+9d
A10=4+9[2]
A 10=4+18
A10=22
An =A1+[n-1]d
An =4[n-1]2
An =4+2n-2
An= 2n+4-2
An=2n+2
5.
A2= 3 x Ab
D= -4
A1=?
An=?
Az = 3 x A6
A 1 +d = 3 x A1+5d
A1+d =3A1+15d
2A1+14d=0
d= -4
2A1+14[-4]=0
2A1+-56= 0
2A1= 56
A1= 28
An = A1+[n-1]d
An= 28+[n-1]-4
An=32-4n
An =32-4n
6.
(a) A
3= 0
d= -2
From the formula
A1+2d= 0
A1+2[-2]= 0
A1-4=0
A1= 4
The first term is 4.

b) The general term
An =A1+ [n-1] d
A n =4+[n-1]-2
An = 4+ -2n+2
An = 6-2n
The general term is 6-2n.

7.
A54 =?
100, 97, 94 = A1, A2, A3

A54= A1+53d
d= A2-A1= A3-A2
d= 97-100= 94-97
d= -3
A54= 100+53[-3]
A 54=100 + -159
A54= -59
8.
4, x, y, 20
A 4= 20
A1 +3d=20 , but A1=4
4+3d= 20
3d=16
d = 16/3
A 1, A2, A3, A4
A2 =A1+d
4+WTPyGSzUw Z3JS7MoejWFxOneEc MgUjmb1xRp0UQWhqiFzeL72DT Z3pq6tmn PM0RO 1HczZc99b1M5ZXOHvuVOENdq7eNzbh9riX1cEI 8zM7b5TP6mZY7apFZ1JA9TDqF I
X = Ci4KsKWgcCm3LVl1t42a K93p4I91YpwXnvUmYEPBZEZYs6G Pbevrv0pSE1m 5LUx171TOZVF8hl KeXZ7kItu5ukSt8t9KGajIixEbYwYoV0zohr9q21qQIB81 Bx IJmNuds
A3 = A1+2d
4+ 2x Ht6T VHTND0dcArf3ld GKq4vb0eRbtijQQIWPEPV7oghQQda9ccut4jXjfVsisbhm1uZyfqwPFWNN1rW6UIpxQcdHGMWYuSrB4 ThnHOmc3c7NsWbsuLtsL6Sl4BjNOJU P1jc
A3 = DYANjYq4c5BT MBqZSNjTUQhAMmoqN3 0QFSuGhW1q RsEqKcuS9E41gEfaq1EmMiE2jdugARNfHDn7sLvp1f7rpPuLzxePXa5lmybcdJMXDQLiqlO8HvgzzjAHm23XTQvSSA
Hence x=Ci4KsKWgcCm3LVl1t42a K93p4I91YpwXnvUmYEPBZEZYs6G Pbevrv0pSE1m 5LUx171TOZVF8hl KeXZ7kItu5ukSt8t9KGajIixEbYwYoV0zohr9q21qQIB81 Bx IJmNuds and y=DYANjYq4c5BT MBqZSNjTUQhAMmoqN3 0QFSuGhW1q RsEqKcuS9E41gEfaq1EmMiE2jdugARNfHDn7sLvp1f7rpPuLzxePXa5lmybcdJMXDQLiqlO8HvgzzjAHm23XTQvSSA
9.
A40 =?
A1 =4
A2= 7
A 3= 10
A1+39d=?
d= 7-4= 10-7
d= 3
A1 +39[3]
A1+117
4+117
A40 =121

10.
A1 =4
A2=9
A3=14
d= 9-4= 14-9
d= 5
An =A1+[n-1]d
An =4+[n-1]5
An =4+5n-5
An =5n-1
The nth term is 5n-1.
11.
a) the common difference
A5= 40
A7= 20
A1+ 4d= 40 ………… (1)
A1+ 6d= 20 ………….(2)
Subtracting equation (2) from equation (1) we obtain
-2d=20
d= -10
b) the tenth term
A10= A1+9d,
But A1 +4d=40
A1=80
A10=A1+9d
=80-90
A 10=-10
12.
A2= A1+d
A7= A1+6d
A1 +6d = 10
A1+d=7
Solving the simultaneous equations by using the elimination method;
-5d= -3
d = 3/5
A1 +KqctiCRFcJQOt8jQ WNUTpWqj7jDEvEko8eb0OFSQ2AXOSpQFPMAzGP2m5xJJ8qSgaYyE18HXuxJs1XMub3KruwiGJx R1k2bpFJ6La Ah4diVmmHRUFP0lSC5LRHtqBVhY P8 = 7
A1 =C4gECeW9LtYfbY65g7fTKhhyzC46M FRSG4nENM7F 3CgvGsj0G0vvUHzYV ZSWK5FTQNmHI48ScBzv F5mi 90mjQBTSI Yaj2lRxbLBVgGnh1oYXhrjj0hFdWwgkh 9FkmAoQ
SUM OF THE FIRST n TERMS OF AN ARITHMETIC PROGRESSION
Consider a series with first term A1, common difference d. If the sum n terms is denoted by Sn, then

Sn = A1 + (A1+ d) + (A1 + 2d)+ …+ (A1 – d) + An
+ Sn = An + (An- d) + (An – 2d)+ …+ (A1 + d) + A1


2Sn = (A1+ An) + (A1+ An) + (A1+ An)+ …+ (A1+ An)+ (A1+ An)

There are n terms of (A1+ An) then
2Sn = n(A1+ An)
Sn = n(A1+ An)
2
... The sum of the first n terms of an A.P with first term A1 and the last termAn is given by
Sn = TDp DH0d CIgI8OwYmdgU7EYVNyd2uWUqmZir4WlVq26BamNF7qGm9WNVOYnxchLhhPFOJ1EF5n70rNP44EStN6e2IdL LIEJislvLZklnaYP9QVum1L8CkApIpQXdAdZAV6vJY(A1+ An)

But An =A1+ (n-1) d
Thus, from
Sn = TDp DH0d CIgI8OwYmdgU7EYVNyd2uWUqmZir4WlVq26BamNF7qGm9WNVOYnxchLhhPFOJ1EF5n70rNP44EStN6e2IdL LIEJislvLZklnaYP9QVum1L8CkApIpQXdAdZAV6vJY(A1+ An)
Sn = TDp DH0d CIgI8OwYmdgU7EYVNyd2uWUqmZir4WlVq26BamNF7qGm9WNVOYnxchLhhPFOJ1EF5n70rNP44EStN6e2IdL LIEJislvLZklnaYP9QVum1L8CkApIpQXdAdZAV6vJY[A1+ A1+ (n-1) d]
Sn = TDp DH0d CIgI8OwYmdgU7EYVNyd2uWUqmZir4WlVq26BamNF7qGm9WNVOYnxchLhhPFOJ1EF5n70rNP44EStN6e2IdL LIEJislvLZklnaYP9QVum1L8CkApIpQXdAdZAV6vJY[2A1+ (n-1) d]

... therefore, the sum of the first n term of an A.P with the first A1 and the common difference d in given by
Sn = TDp DH0d CIgI8OwYmdgU7EYVNyd2uWUqmZir4WlVq26BamNF7qGm9WNVOYnxchLhhPFOJ1EF5n70rNP44EStN6e2IdL LIEJislvLZklnaYP9QVum1L8CkApIpQXdAdZAV6vJY[2A1+ (n-1) d]
Where
n =number of terms
A1= first term
An =last term
d= common difference
Example
i) Find sum of the first 5th term where series is 2, 5, 8, 11, 14 first formula
Solution
S5 =C5XexQEcqcbT4vOMOExrCHk RXAb7kjKa2XR6Y9CztEOVRdEpfKW4HPiYK8D EbGx7w3qlyEPx6YXObiJ5NGQGRj52lzbr7nyVn0UtcQifWg YEA1brDUWj1122Yzj8Hr0zxjFs [2 + 14]
S5 = 40
(ii)Sn =TDp DH0d CIgI8OwYmdgU7EYVNyd2uWUqmZir4WlVq26BamNF7qGm9WNVOYnxchLhhPFOJ1EF5n70rNP44EStN6e2IdL LIEJislvLZklnaYP9QVum1L8CkApIpQXdAdZAV6vJY [2A1 + [n-1] d]
S5 =C5XexQEcqcbT4vOMOExrCHk RXAb7kjKa2XR6Y9CztEOVRdEpfKW4HPiYK8D EbGx7w3qlyEPx6YXObiJ5NGQGRj52lzbr7nyVn0UtcQifWg YEA1brDUWj1122Yzj8Hr0zxjFs [2 x 2 + [5-1] (3)]
S5= 40

Arithmetic Mean

If a, m and b are three consecutive terms of an arithmetic. The common difference
d= M – a
Therefore M- a = b – M
2M = a+ b
M= a + b
2
M is called the arithmetic mean of and b
E.g. Find the arithmetic mean of 3 and 27
M = 3+ 27
2
= 30 = 15
2

GEOMETRIC PROGRESSION (G.P)
Defination:
Geometric progression is a series in which each term after the first is obtained by multiplying the preceding term by the fixed number.
The fixed number is called the common ratio denoted by r.
E.g. 1+2+4+8+16+32+…
3+6+12+24+48

The nth term of Geometric Progression
If n is the number of terms of G.P, the nth term is denoted by Gn and common ratio by r. Then G n+1 = rGn for all natural numbers.
G1 = G1
G2 = G1r
G3 = G2r = G1r.r = G1r2
G4 = G3r = G1r2.r = G1r3
G5 = G4r = G1r3.r = G1r4
G6 = G5r = G1r4.r = G1r5
G7 = G6r = G1r5.r = G1r6
G8 = G7r = G1r6.r = G1r7
The nth term is given by
Gn = G1rn-1

Example1: Write down the eighth term of each of the following.
(a). 2+ 4+ 8+…
(b). 12+ 6+ 3+ …
Solution
(a) The first term G1 =2, the common ratio r=2 and n=8, Then from
Gn = G1rn-1
G8 =(2)(2)8-1
G8 = (2). 27
G8 = 256
(b) The first term G1 =12, the common ratio r=1/2 and n=8, Then from
Gn = G1rn-1
G8 =(12)(1/2)8-1
G8 = (12). (1/2)7
G8 = 12/128 = 3/32

Example2: Find the numbers of terms in the following 1+2+4+8+16+…+512.

Solution
The first term G1 =1, the common ratio r=2 and Gn= 512, Then from
Gn = G1rn-1
512 =(1)(2)n-1
512 = 2n-1
512 = 2n.2-1
512 x 2= 2n
1024= 2n
210 = 2n
n = 10

The sum of the first n terms of a geometrical progression.

Let the sum of first n terms of a G.P be denote by Sn
Sn = G1+ G2+ G3+ G4+…+ Gn
Since the common ratio is r
From
G2 = G1r
G3 = G1r2
G4 = G1r3
…Gn = G1rn-1
Sn = G1+ G1r+ G1r2+ G1r3+…+ Gnrn-1
Multiplying by common ratio r both sides we have

rSn