Question obtain the value of X for which (X+1),(X-5),(X-2) is a geometric progression.hence find the sum of the first 12 terms of the progression.

If x + 1, x – 5, and x – 2 are in a geometric progression, then there is some constant r for which x – 5 = r (x + 1) ==> r = (x – 5) / (x + 1) and x – 2 = r (x – 5) ==> r = (x – 2) / (x – 5) Then (x – 5) / (x + 1) = (x – 2) / (x – 5) Solve for x : (x – 5)² = (x – 2) (x + 1) x ² – 10x + 25 = x ² – x – 2 -9x = -27 x = 3 It follows that the ratio between terms is r = (3 – 5) / (3 + 1) = -2/4 = -1/2 Now, assuming x + 1 = 4 is the first term of the G.P., the n-th term a(n) is given by a(n) = 4 (-1/2)ⁿ⁻¹ The sum of the first 12 terms – denoted here by S – is then S = 4 (-1/2)⁰ + 4 (-1/2)¹ + 4 (-1/2)² + … + 4 (-1/2)¹¹ Solve for S : S = 4 [(-1/2)⁰ + (-1/2)¹ + (-1/2)² + … + (-1/2)¹¹] (-1/2) S = 4 [(-1/2)¹ + (-1/2)² + (-1/2)³ + … + (-1/2)¹²] ==> S – (-1/2) S = 4 [(-1/2)⁰ – (-1/2)¹²] ==> 3/2 S = 4 (1 – 1/4096) ==> S = 8/3 (1 – 1/4096) ==> S = 1365/512 Log in to Reply

If

x+ 1,x– 5, andx– 2 are in a geometric progression, then there is some constantrfor whichx– 5 =r(x+ 1)==>

r= (x– 5) / (x+ 1)and

x– 2 =r(x– 5)==>

r= (x– 2) / (x– 5)Then

(

x– 5) / (x+ 1) = (x– 2) / (x– 5)Solve for

x:(

x– 5)² = (x– 2) (x+ 1)x² – 10x+ 25 =x² –x– 2-9

x= -27x= 3It follows that the ratio between terms is

r= (3 – 5) / (3 + 1) = -2/4 = -1/2Now, assuming

x+ 1 = 4 is the first term of the G.P., then-th terma(n)is given bya(n)= 4 (-1/2)ⁿ⁻¹The sum of the first 12 terms – denoted here by

S– is thenS= 4 (-1/2)⁰ + 4 (-1/2)¹ + 4 (-1/2)² + … + 4 (-1/2)¹¹Solve for

S:S= 4 [(-1/2)⁰ + (-1/2)¹ + (-1/2)² + … + (-1/2)¹¹](-1/2)

S= 4 [(-1/2)¹ + (-1/2)² + (-1/2)³ + … + (-1/2)¹²]==>

S– (-1/2)S= 4 [(-1/2)⁰ – (-1/2)¹²]==> 3/2

S= 4 (1 – 1/4096)==>

S= 8/3 (1 – 1/4096)==>

S=1365/512