Given that 2m – 8, 2m + 4 and 5m – 2 are the first three successive terms of a geometric sequence. Find the value of m and thus the summation of the first 10 e

Question

maths and algorithms

Answer 1

In a geometric sequence, the ratios of successive terms are equal:  a₂/a₁ = a₃/a₂ = ... = a_n+1/a_n.  In your sequence, the successive terms are:

 

a₁ = 2m-8

a₂ = 2m+4

a₃ = 5m-2

 

a₂/a₁ = a₃/a₂

(2m+4)/(2m-8) = (5m-2)/(2m+4)

(2m+4)² = (5m-2)(2m-8)

4m² + 16m + 16 = 10m² - 44m + 16

0 = 6m² - 60m

0 = 6m(m-10)

 

Solve for m.  You'll get two values.  They both work but produce separate sequences.  The sum of the first n terms of each sequence is:

 

S = a1(1-rⁿ)/(1-r)    for r≠1

• S = the sum of the first n terms
• n = number of terms = 10
• r = common ratio = ratio of successive terms in each sequence (a_n+1/a_n)