Question

The sum of the first 10 terms of an AP is four times the sum of the first 5 terms then the ratio of the first term to the common difference is

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Answer 1

The ratio of the first term to the common difference is 1 : 2

Explanation:

The rule of the sum of the AP is

`S_(n)=(n)/(2)[2a+(n-1)d]` , where

a is the first number

d is the common difference

n is the position of the number

∵
`S_(10)` = 4 ×
`S_(5)` ⇒ (1)

→ Find
`S_(10)` and
`S_(5)`

∵ In
`S_(10)` n = 10

∴
`S_(10)=(10)/(2)[2a+(10-1)d]`

∴
`S_(10)=5[2a+9d]`

∴
`S_(10)` = 10a + 45d ⇒ (2)

∵ In
`S_(5)` n = 5

∴
`S_(5)=(5)/(2)[2a+(5-1)d]`

∴
`S_(5)=2.5[2a+4d]`

∴
`S_(5)` = 5a + 10d ⇒ (3)

→ Substitute (2) and (3) in (1)

∵ 10a + 45d = 4[5a + 10d]

∴ 10a +45d = 20a + 40d

→ Subtract 40d from both sides

∴ 10a + 45d - 40d = 20a + 40d - 40d

∴ 10a + 5d = 20a

→ Subtract 10a from both sides

∴ 10a - 10a + 5d = 20a - 10a

∴ 5d = 10a

→ Divide both sides by 5

∴ d = 2a

→ That means d is twice a or a is half d

∴ a : d = 1 : 2

The ratio of the first term to the common difference is 1 : 2