What is the 26th term of the geometric sequence with a5 = 5/4 and a12 = 160?

Let r be the common ratio, then

T(5)=T(1)r^4=5/4-------------(1)

T(12)=T(1)r^11=160-----------(2)

(2)/(1)=>

r^7=160(4)/5=>

r=2

From (1), get

T(1)=5/(4*16)=>

T(1)=5/64

T(26)=(5/64)(2^25)=>

T(26)=2621440

For a geometric progression tn=ar^n-1. T5=ar^4=5/4.......(1). T12=ar^11=160.....(2). (2)/(1)= r^7=128=2^7. r=2. Substitute for r in 1. a2^4=5/4. a*16=5/4. a=5/4/16=5/64. T26=ar^25=5/64 * 2^25=5*2^25/64=5*2^25/2^6=5*2^(25-6)=5*2…

first let a5 =a1=5/4and a12 = a8=160

used the formula an = a1r^n-1

a8 = a1r^8-1,substitute the given values

160 =5/4r^8-1

160 = 5/4r^7,divide both sides by 5/4

128 = r^7,raised to the power of 1/7

2=r,so we can now solve for a1 using a5=a1r^5-1

5/4=a1(2)^4

5/4=a1(16),divide both sides by 16

5/64=a1

let n = 26, a1=

a26 = a1r^26-1

a26 = 5/64(2)^25

a26=2621440,answer

a(n) = a(1) * r^(n - 1)

Therefore:

a(5) = a(1) * r^(5 - 1) = a(1) * r^4 = 5/4

a(12) = a(1) * r^(12 - 1) = a(1) * r^11 = 160

Divide those last two equations:

r^7 = 160 / (5/4)

r^7 = 160 * (4/5)

r^7 = 128

r = 128^(1/7)

r = 2

Substitute that into the first equation:

a(1) * 2^4 = 5/4

a(1) * 16 = 5/4

a(1) = (5/4) / 16

a(1) = 5/64

Therefore:

a(26)

= a(1) * r^(26 - 1)

= a(1) * r^25

= (5/64) * 2^25

= (5/64) * 33554432

= 2621440