Question

The sum of three numbers of a GP is 26 and their product is 216 .find the numbers

Answer 1

let three numbers of G.P are , a ,ar ,ar^2

given, sum = 26

a + ar + ar^2 = 26. --------------------------(1)

a + ar + ar . r = 26. ----------------- -------(2)

product = 216

a × ar× ar^2 = 216

a^3 r ^3 = (6)^3

( ar )^3 = 6 ^3

ar = 6 ----------------------------------------(3)

put value of" ar " in eq.(2) , we get

a + 6 + 6.r = 26

a + 6r = 20

a = 20 - 6r -----------------------------------(4)

put value of "a" in eq. (3) , we get

( 20 - 6r ) r = 6

20r - 6r^2 = 6

6r^2 - 20r + 6 = 0

2( 3r^2 - 10r + 3 ) =0

3r^2 - 10r + 3 = 0

3r^2 - 9r - r + 3 =0

3r( r - 3 ) - 1 ( r - 3 )

( 3r - 1 )( r - 3 )=0

r = 3 , 1 / 3

put integer value of r = 3 in eq.(4) ,we get,

a = 20 - 6(3) = 2

a = 2

three numbers are , a = 2

ar = 2 × 3 = 6

ar^2 = 2 × ( 3 )^2 = 18

hence three numbers are , 2, 6, 18 whose sum (2 + 6+ 18 ) = 26 and product ( 2×6 ×18) = 216

【 hope it helps】 【mukeshbhardwaj】

Answer 2

Let the terms be a ,ar,ar2. Then we are given.                                                            a+ar+ar3 =26_______1.                              a3r3 = 216__________2.                   from eq 2  ar = 6_______3        substitute  eq 3 in 1                                      a+ 6 r = 29 __________4.                             from eq 3 ,r= 6/a_________5.                     Substitute eq 5 into 4                                     a2- 20a +36=0.                                               (a-2) (a-18) = 0.                                            a=2,r=3 (or) a=18,r=1/3.                                  So the three terms are 2,6 (or) 18,6,2