Question

If the sum of three consecutive numbers in a geometric progression is 26 and the sum of their squares is 364, then the product of those numbers is
(1) 216
(2) 125
(3) 343
(4) 512

Answer 1

Answer:

2, 6 and 18

Step-by-step explanation:

Let the three consecutive terms in GP be x,y,z

Then, y² = xz............(1)


As per given data,

x + y + z=26...........(2)

x² +y² +z² =364......(3)


we know that

(x+y+z)² = x² + y² + z² +2(xy+yz+zx)

using equation (1)

(x+y+z)² = x² + y² + z² +2(xy+yz+y²)

(x+y+z)² = x² + y² + z² +2(x+y+z)y

(26)² = 364 + 2(26)y

676 - 364 = 52y

312 = 52 y

y=6

Now equations (1) and (2) becomes

xz= 36, x+z=20

By trial and error method we can easily find out

x = 2 and z = 18

Therefore the required three numbers are

x =2, y =6, z =18

Then, xyz= (2)(6)(18)=216



:) oR :(

Let three consecutive terms in geometric progression are : a/r , a , ar

a/c to question,

a/r + a + ar = 26

or, a[1/r + 1 + r] = 26

a(1 + r + r²)/r = 26

squaring both sides,

a²(1 + r + r²)²/r² = 676 .........(1)

again, a²/r² + a² + a²r² = 364

or, a²[1/r² + 1 + r² ] = 364

or, a² (1 + r² + r⁴)/r² = 364

a²(1 + r - r²)(1 + r + r²)/r² = 364.......(2)


from equations (1) and (2),

(1 - r + r²)/(1 + r + r²) =364/676

(1 - r + r²)/(1 + r + r²) = 14/26 = 7/13

13(1 - r + r²) = 7(1 + r + r²)

13 - 13r + 13r² = 7 + 7r + 7r²

6 - 20r + 6r² = 0

3r² - 10r + 3 = 0

3r² - 9r - r + 3 = 0

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

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

r = 3 and 1/3

put r = 3 or 1/3 in equation (1),

a(1 + 3 + 3²)/3 = 26

a × 13/3 = 26

a = 6

hence, product of a/r, a , ar = a³ = (6)³ =216