Question

(ii) Find the A.P. whose 1st term is 100 and the sum of the first six terms is 5 times
the sum of the next six terms.​

Answer 1

Answer:

100,90,80,70,60,50,40,30,20

Step-by-step explanation:

refer to attachment

Answer 2

Answer:

The required AP is 100, 90, 80, 70, ....

Step-by-step-explanation:

We have given that,

The first term of an AP is 100.

∴ a = t₁ = 100

Now, we know that,

Sₙ = ( n / 2 ) [ 2a + ( n - 1 ) * d ] - - - [ Formula ]

Now, the sum of the first six terms of the AP,

S₆ = ( 6 / 2 ) [ 2a + ( 6 - 1 ) * d ]

⇒ S₆ = ( 6 / 2 ) [ 2a + 5d ]

⇒ S₆ = ( 6 / 2 ) ( 2a + 5d )

⇒ S₆ = 3 ( 2a + 5d )

⇒ S₆ = 6a + 15d

⇒ S₆ = 6 * 100 + 15d - - - [ Given ]

⇒ S₆ = 600 + 15d

⇒ S₆ = 15d + 600 - - - ( 1 )

Now, the sum of the first twelve terms of the AP,

S₁₂ = ( S₆ + S₆ⁿᵉˣᵗ )

⇒ S₆ⁿᵉˣᵗ = S₁₂ - S₆

⇒ S₆ⁿᵉˣᵗ = ( 12 / 2 ) [ 2a + ( 12 - 1 ) * d ] - ( 15d + 600 ) - - - [ From ( 1 ) ]

⇒ S₆ⁿᵉˣᵗ = 6 ( 2a + 11d ) - 15d - 600

⇒ S₆ⁿᵉˣᵗ = 12a + 66d - 600 - 15d

⇒ S₆ⁿᵉˣᵗ = 12a + 66d - 15d - 600

⇒ S₆ⁿᵉˣᵗ = 12a + 51d - 600

⇒ S₆ⁿᵉˣᵗ = 12 * 100 + 51d - 600

⇒ S₆ⁿᵉˣᵗ = 1200 + 51d - 600

⇒ S₆ⁿᵉˣᵗ = 51d + 1200 - 600

⇒ S₆ⁿᵉˣᵗ = 51d + 600 - - - ( 2 )

Now, from the given condition,

S₆ = 5 ( S₆ⁿᵉˣᵗ )

⇒ 15d + 600 = 5 ( 51d + 600 ) - - - [ From ( 1 ) & ( 2 ) ]

⇒ 15d + 600 = 255d + 3000

⇒ 255d - 15d = 600 - 3000

⇒ 240d = - 2400

⇒ d = - 2400 ÷ 240

⇒ d = - 10

Now,

a = t₁ = 100

d = - 10

∴ t₂ = t₁ + d

⇒ t₂ = 100 + ( - 10 )

⇒ t₂ = 100 - 10

⇒ t₂ = 90

Now,

t₃ = t₂ + d

⇒ t₃ = 90 + ( - 10 )

⇒ t₃ = 90 - 10

⇒ t₃ = 80

Now,

t₄ = t₃ + d

⇒ t₄ = 80 + ( - 10 )

⇒ t₄ = 80 - 10

⇒ t₄ = 70

∴ The required AP is 100, 90, 80, 70, ....