Sequences & Series

Edexcel IAL WMA12/01/P2/June 2022/Q6 (Geometric Series, Sum to infinity)

In a geometric sequence​​ u1 , u2 , u3 , … 

• the common ratio is r​​ 

• u2 + u3 = 6​​ 

• u4 = 8 

• Show that r satisfies​​ 

3r2 - 4r - 4 = 0

(3)​​ 

Given that the geometric sequence has a sum to infinity,​​ 

• find​​ u1​​ 

(3)​​ 

• find​​ S∞​​ 

(2)

SOLUTION​​ 

a- Since its is geometric sequence, it has a commeon ratio.​​ 

u2u1=u3u2=u4u3=r 

u2u1=u3u2=8u3=r 

Since​​ u2+u3=6, so substituting ​​ u2=6-u3. Moreover,​​ u4 = 8, so the expressions on substituting both becomes

6-u3u1=u36-u3=8u3=r 

u32=86-u3    ….(1)

Now, as​​ 8u3=r, so we may rewrite the expression in terms of​​ u3​​ as​​ 

u3=8r

Putting back in equation 1,​​ 

8r2=86-8r

64r2=48r2-64r

48r2-64r-64=0 

Hence, the simplified equations comes out to be

3r2-4r-4=0  

b- To find​​ u1, which is the basically the first term of the sequence, also known as ‘a’,​​ we may use the given data that​​ u4=8. But before that let us find the value of r by solving the quadratic equation we got in part (a).​​ 

3r2-4r-4=0  

3r+2r-2=0

3r=-2             r=2

r=-23             r=2

As it is mentioned in the question that the geometric sequence has a sum to infinity so then​​ r<1. Therefore, we will be choosing ​​ r=-23.

So, now putting r value in​​ 

un=arn-1

u4=a×-234-1

8=a×-827

a=27 x 8-8

a=a= -27

u1=-27

c- To find the sum to infinity of a convergent geometric series, use the following formula.​​ 

S∞=a1-r

S∞=-271--23=-2753

S∞= -815