Question

In a geometric progression the second term is nine less than the first term. The sum of the second and third terms is 30. Given that all the terms of the progression are positive find the first term.

Answer 1

N general, if a GP has a 1st term a, and a common ratio r, then the sequence of terms is : a, ar, ar², ar³, . . ., arⁿ⁻¹. 
Then, a = ar + 9 . . .so a - ar = 9 : a(1 - r) = 9 : a = 9/(1 - r) 
and ar + ar² = 30 . . . so a(r + r²) = 30 : a = 30/(r + r²) . . equating the two expressions for a gives :- 
9/(1 - r) = 30/(r + r²) : 9r + 9r² = 30 - 30r : 9r² + 39r - 30 = 0 : 3r² + 13r - 10 = 0 . . . this factors as (3r - 2)(r + 5) = 0. 
Then r = ⅔ or r = -5 . . . if we used r = -5 then successive terms would alternate from positive to negative. As all terms are positive then r = ⅔. 
Thus [using the equation a = 9/(1 - r)] : a = 9/⅓ = 27. . . . which is the required answer. 

Check : 2nd term : ar = 27 x ⅔ = 18 which is 9 less than the 1st term 
3rd term : ar² = 27 x (4/9) = 12 . . . and 18 + 12 = 30.