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

Answer:

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

Step-by-step explanation:

The rule of the sum of the AP is

$S_{n}=\frac{n}{2}[2a+(n-1)d]$ , 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

∴ $S_{10}=\frac{10}{2}[2a+(10-1)d]$

∴ $S_{10}=5[2a+9d]$

∴ $S_{10}$ = 10a + 45d ⇒ (2)

∵ In $S_{5}$ n = 5

∴ $S_{5}=\frac{5}{2}[2a+(5-1)d]$

∴ $S_{5}=2.5[2a+4d]$

∴ $S_{5}$ = 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