Question

Q2.The sum of four numbers is G.P.is 60 and the A.M.between first and the last is 18 .Find the numbers.​

Answer 1

Answer:

32, 16, 8, 4 or 4, 8, 16, 32

Step-by-step explanation:

Let the 1st, 2nd, 3rd, 4th number be a, ar , ar² and ar³ respectively.

AM b/w a and ar³ = 18

=> (a + ar³)/2 =18

=> a + ar³ = 36 ... (1)

=> a(1 + r³) = 36 ...(2)

Sum of all numbers = 60

=> a + ar + ar² + ar³ = 60

=> a + ar³ + ar + ar² = 60

=> 36 + ar + ar² = 60 {from (1)}

=> a(r + r²) = 24 ... (3)

After dividing (2) by (3):

=> (1 + r³)/(r + r²) = 3/2

=> 2 +2r³ = 3r + 3r²

=> 2r² - 3r² - 3r + 2 = 0

=> (r + 1)(2r - 1)(r - 2) = 0

=> r = - 1, r = 1/2 , r = 2

Based on these values, put r in (1),

a = 36/0 (not defined, so neglect r = -1)

a = 32, r = 1/2

a = 4 , r = 2

Terms are,

32, 16, 8, 4 or 4, 8, 16, 32

Answer 2

Answer:

• $\sf{32,\:16,\:8,\:4 \:\:Or \:\: 4\:, 8,\: 16, \:32 }$

Step-by-step explanation:

Let the four terms of GP be a , ar , ar² , ar³.

According to the first statement,

The sum of four numbers in GP is 60.

→ a + ar + ar² + ar³ = 60 ____eqn.(1)

According to the second statement,

The A.M.between first and the last is 18.

→ a + ar³/2 = 18

→ a + ar³ = 36 ________eqn.(2)

Substitute the value of eqn.(2) in eqn.1

→ a + ar³ + ar+ ar² + = 60

→ 36 + ar + ar² = 60

→ ar + ar² = 24 ________eqn.(3)

Divide eqn.(2) & (3)

$\sf \longrightarrow \: \frac{a + ar {}^{3} }{ar + ar {}^{2} } = \frac{36}{24} \\ \\ \sf\longrightarrow \: \frac{1 + r {}^{3} }{r+ r {}^{2} } = \frac{3}{2} \\ \\ \sf\longrightarrow \: 2( 1 + r {}^{3} ) = 3(r + r {}^{3} )\\ \\ \sf\longrightarrow \: 2r {}^{3} - 3r {}^{2} - 3r + 2 = 0 \\ \\ \longrightarrow \: \boxed{\sf \: \red{r = - 1, \frac{1}{2}, 2 }} \: \: \green\bigstar$

Put r = 1/2 in eqn. (2)

$\sf \longrightarrow \: a(1 + \frac{1}{2} {}^{3} ) = 36 \\ \\ \sf\longrightarrow \: \green{a = 32} \\ \\ \therefore \: \boxed{\sf a</p><p></p><p>,</p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p> ar,ar {}^{2} ,ar {}^{3} = 32,16,8,4 } \: \pink \bigstar$

Put r = 2 in eqn.(2)

$\sf \longrightarrow \: a(1 + 2{}^{3} ) = 36\\ \\ \sf\longrightarrow \: \pink{a = 4} \\ \\ \therefore \: \boxed{\sf a</p><p></p><p>,</p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p></p><p> ar,ar {}^{2} ,ar {}^{3} = 4</p><p></p><p></p><p>,</p><p></p><p>8</p><p></p><p>,</p><p></p><p>16</p><p></p><p>,</p><p></p><p>32} \: \purple \bigstar$