Sum of Geometric Series

Sum of Geometric Series

 

The series associated with a geometric sequence is known as a geometric series. For example, 2 + 4 + 8 + 16 + 32 is the geometric series associated with the geometric sequence 2, 4, 8, 16, 32.


 

Let a be the first term, r be the common ratio, n be the number of terms, l be the last term and Sn be the sum to n terms of GS, then

 

Sn = a + ar + ar2 + … … … + arn-2 + arn-1 ………. (i)

 

Multiplying both sides by r,

 

rSn = ar + ar2 + ar3 + … … … + arn-1 + arn …….. (ii)

 

Subtracting (i) from (ii), we get

 

(r – 1)Sn = -a + arn


∴ Sn = a(r^n-1)/(r-1)  Again,  Sn = (ar^n-a)/(r-1) or, Sn = (ar^(n-1).r-a)/(r-1) or, Sn = (lr-a)/(r-1)    [∵ l = arn-1]  ∴ Sn = (lr-a)/(r-1)


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Worked Out Example

 

Example 1: Find the sum of the geometric series 1 + 2 + 4 + … … … 7 terms.

 

Solution:

 

Here,

 

First term (a) = 1

Common ratio (r) = 2/1 = 2

Number of terms (n) = 7

Sum of the terms (Sn) = ?

 

By using formula,

 

Sn = a(r^n-1)/(r-1) or, Sn = 1(2^7-1)/(2-1) or, Sn = (128-1)/(2-1) or, Sn = 127/1 or, Sn = 127

Sn = 127


 

Example 2: In a GP, the first term is 7, the last term is 448 and the sum is 889, find the common ratio.

 

Solution:

 

Here,

 

First term (a) = 7

Last term (l) = 448

Sum of terms (Sn) = 889

Common ratio (r) = ?

 

By using the formula,

 

Sn = (lr-a)/(r-1) or, 889 = (448r-7)/(r-1) or, 889r – 889 = 448r – 7 or, 889r – 448r = 889 – 7 or, 441r = 882 or, r = 882/441 or, r = 2

The common ratio is 2.

 

 

Example 3: Find the sum of ∑_(n=1)^5▒3^(n+1)

Solution:

 

Here,

 

∑_(n=1)^5▒3^(n+1)
= 31+1 + 32+1 + 33+1 + 34+1 + 35+1

= 32 + 33 + 34 + 35 + 36

= 9 + 27 + 81 + 243 + 729

= 1089

 

 

Example 4: How many terms of the series 32 + 48 + 72 + … … will add upto 665?

 

Solution:

 

Here,

 

First term (a) = 32

Common ratio (r) = 48/32 = 3/2

Sum of terms (Sn) = 665

Number of terms (n) = ?

 

By using formula,


Sn = a(r^n-1)/(r-1) or, 665 = 32{(3/2)^n-1}/(3/2-1) or, 665 = 32{(3/2)^n-1}/(1/2) or, 665 = 64{(3/2)^n-1} or, 665/64 = (3/2)^n-1 or, 665/64 + 1 = (3/2)^n or, (665+64)/64 = (3/2)^n or, 729/64 = (3/2)^n or, (3/2)^6 = (3/2)^n

n = 6

 

The number of terms = 6

 

 

Example 5: If S3 and S6 of a GS are 7 and 63 respectively, find the common ratio.

 

Solution:

 

Here,

 

S3 = 7 or, a(r^3-1)/(r-1) = 7 ……………. (i)  S6 = 63 or, a(r^6-1)/(r-1) = 63 …………. (ii)  Dividing equation (ii) by (i),  (a(r^6-1)/(r-1))/(a(r^3-1)/(r-1)) = 63/7 or, a(r^3+1)(r^3-1)/(r-1) × (r-1)/a(r^3-1)  = 63/7
or, r3 + 1 = 9

or, r3 = 9 – 1

or, r3 = 8

or, r3 = 23

or, r = 2

 

Common ratio = 2

 

 

Example 6: The second and fifth terms of GS are 3 and 81 respectively. Find the sum of the first five terms.

 

Solution:

 

Here,

 

The second term of GS is 3

i.e. t2 = 3

or, ar = 3

 

Again, the fifth term of GS is 81

i.e. t5 = 81

or, ar4 = 81

or, ar.r3 = 81

or, 3.r3 = 81

or, r3 = 81/3

or, r3 = 27 = 33

or, r = 3

 

And, ar = 3

or, a×3 = 3

or, a = 1

 

Now,

S5 = a(r^5-1)/(r-1) = 1(3^5-1)/(3-1) = (243-1)/2 = 242/2 = 121.

 

Example 7: Find the GP for which the sum of the first two terms is -4 and the fifth term is 4 times the third term.

 

Solution:

 

Here,

 

S2 = -4

i.e. t1 + t2 = -4

or, a + ar = -4

or, a(1 + r) = -4 ……………… (i)

 

t5 = 4 × t3

i.e. ar4 = 4 × ar2

or, r2 = 4

or, r = ±2

 

Taking r = 2, from equation (i), we have

a(1 + 2) = -4

or, a = -4/3

 

The GP is a, ar, ar2, …

i.e. -4/3, -4×2/3, -4×22/3, …

i.e. -4/3, -8/3, -16/3, …

 

Taking r = -2, from equation (i), we have

a(1 – 2) = -4

or, a = 4

 

The GP is a, ar, ar2, …

i.e. 4, 4×-2, 4×(-2)2, …

i.e. 4, -8, 16, …


 

Example 8: If the sum of the first three terms of a GP is 1 and the sum of the first six terms is 28. Find the sum of the first 9 terms of the series.

 

Solution:

 

Here,

 

S3 = 1 or, a(r^3-1)/(r-1) = 1 ……………. (i)  S6 = 28 or, a(r^6-1)/(r-1) = 28 …………. (ii)  Dividing equation (ii) by (i),  (a(r^6-1)/(r-1))/(a(r^3-1)/(r-1)) = 28/1 or, a(r^3+1)(r^3-1)/(r-1) × (r-1)/a(r^3-1)  = 28
or, r3 + 1 = 28

or, r3 = 28 – 1

or, r3 = 27

or, r3 = 33

or, r = 3

 

Substituting r = 3 in (i), a(3^3-1)/(3-1) = 1 or, a(27 – 1) = 2 or, a = 1/13  Now, the sum of the first 9 terms, Sn = a(r^n-1)/(r-1) or, S9 = (1/13(3^9-1))/(3-1) or, S9 = (19683-1)/26 or, S9 = 757
 

The sum of first 9 terms is 757.



 

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