Sequences & Series

Edexcel IAL WMA12/01/P2/Oct 2021/Q7 (Geometric Sequences & Series)

• A geometric sequence has first term 4 and common ratio 6​​ 

Given that the nth term is greater than​​ 10100, find the minimum possible value of n.​​ 

(3)​​ 

• A different geometric sequence has first term a and common ratio r.​​ 

Given that​​ 

• the second term of the sequence is −6​​ 

• the sum to infinity of the series is 25​​ 

• show that​​ 

25r2 - 25r - 6 = 0 

(3)​​ 

• Write down the solutions of​​ 

25r2 - 25r - 6 = 0 

(1)

Hence,​​ 

• state the value of r, giving a reason for your answer,​​ 

(1)​​ 

• find the sum of the first 4 terms of the series.​​ 

(2)

SOLUTION​​ 

i- It is given that​​ a=4,​​ ​​ r=6, and​​ un=10100; applying the geometric nth term formula to find the minimum possible value of n, which is greater than​​ 10100. 

un>10100

arn-1>10100

46n-1>10100

6n-1>101004

n-1>log6⁡101004

n-1>log6⁡10100-log6⁡4

n-1>100log6⁡10 -log6⁡4

n-1>127.73…

n>128.73…

n=129

ii- a- ​​ it is given that the second term of the geometric sequence is −6.​​ 

un=arn-1

u2= -6

ar= -6

a= -6r

Moreover, it is given that the sum to infinity of the series is 25. ​​ 

S∞=25 

a1-r=25

a x11-r=25

Substituting the value of a in terms of r that we just found above .​​ 

-6r x11-r=25

-6r1-r=25

-6=25r1-r

-6=25r-25r2

25r2-25r-6=0 

ii- b- Solving the quadratic equation to get the values of r.​​ 

r=--25±-252-425-6225=

r=65 ,-15

ii- c- As this series has a sum to infinity, this is a convergent geometric series so its​​ r<|.​​ 

Therefore, we will accept​​ 

r=-15

ii- d- To find the sum of the 4 terms of the given series, where​​ r= -15. So let us start by finding a.​​ 

a= -6r

 a= -6-15

a=30

Now, using the forumula of geometric series formula 1 as​​ r<1.

Sn=a1-rn1-r 

S4=30 1--15n1--15 

S4=62425 

S4=24.96