If 6^ th term of a geometric progression is - 1 32 and 9^ th term is 1 256 then r is

Step-by-step solution

Let the first term of the geometric progression be a and the common ratio be r . The n^ th term of a geometric progression is given by T_n = ar^ n-1 . Given T₆ = ar^5 = - 1 32 and T₉ = ar^8 = 1 256 . Dividing the expression for T₉ by T₆ , we get: ar^8 ar^5 = 1/256 -1/32 r^3 = 1 256 (-32) = - 32 256 r^3 = - 1 8 Taking the cube root on both sides, we get r = - 1 2 . Answer: - 1 2