Question

the product of three of a GP 512.if 8 is added to the first term and 6 to the second term find the new term form an AP.

Answer 1

lets take the first term of the G.P. to be a and common ratio to be r. then first three terms of the G.P. => a,ar and ar^2 . if we multiply the three terms ,we get a^3r^3 =512(given) .on applying cube root on both sides we get ar=8. So the other terms can be written as 8/r ,8, 8r . it is said that if we add 8 to first term and 6 to second term it forms a A.P.
First term is 8/r+ 8,Second term is 8+6=14 and third term is 8r . In an A.P. ,the second term when multiply by 2 = sum of first term and third term => 2(14) = 8/r+8+8r => 28-8 =8/r +8r => 20= 8/r +8r
=> multiply each term by r , we get
$20r = 8 + 8 {r}^{2} = > 2 {r}^{2} - 5r + 2 = 0$
=>
$2 {r}^{2} - 4r - r + 2 = 0 = > 2r(r - 2) -$
$1(r - 2) = 0$
=>
$(2r - 1)(r - 2) = 0$
=> so we get r=1/2 or 2 => The terms of the G.P.
are 8/(1/2),8,8(1/2)=> 16,8,4 .Hope it helps you...