Sequences & Series

Edexcel IAL WMA12/01/P2/Jan 2023/Q7 (Geometric Sequences: Summation)

A geometric sequence has first term a and common ratio r, where r > 0​​ 

Given that​​ 

• the 3rd term is 20​​ 

• the 5th term is 12.8​​ 

• show that r = 0.8​​ 

(1)​​ 

• Hence find the value of a.​​ 

(2)​​ 

Given that the sum of the first n terms of this sequence is greater than 156​​ 

• find the smallest possible value of n.​​ 

(Solutions based entirely on graphical or numerical methods are not acceptable.)​​ 

(4)

SOLUTION

a- Using the geometric sequemce formula, to find the value of common ration, r.​​ 

U3=20    ⇒  ar2=20      -1

U5=12.8  ⇒  ar4=128            -2

On dividing equation 2 by 1, we get​​ 

 ar4ar2=12.820⇒r2=12.820.8  

r=12.820 

r= 0.8 

b- To find the first term a, substitute the value of r found in part (a) in any one of the above worked equations.​​ 

(You may also substitite the value of r in the second equation.)

ar2=20

Where,​​ r=0.8. So,​​ 

a=200.82= 1254

a=31.25

c- ​​ To find the smallest possible value of n, use the sum formula.​​ 

Since​​ r<0​​ in our case, we will use the first formula of geometric series.​​ 

Sn=a1-rn1-r

Where,​​ 

a=31.25 

r= 0.8

It is given in the question that the sum of the given geometric series is greater than 156.​​ 

Sn>156

31.25 1-0.8n1-0.8>156

1-0.8n>156 x 0.231.25

1-0.8n>624625

1-624325>0.8n 

1625>0.8n

log10⁡1625>log10⁡0.8n

log10⁡1625>nlog10⁡0.8 

Remember, to swap the inequality sign because if​​ N<1​​ then​​ loga⁡N<0, and in our case (log10⁡0.8), N is 0.8 which is less than 1 therefore​​ log10⁡0.8​​ is less than zero.​​ 

log10⁡1625log10⁡0.8<n

n>28.850

n=29