Question

An arithmetic series has first term a and common difference d.

The sum of the first 2n terms of the series is four times the sum of the first n terms of
the series.

Find an expression for a in terms of d.

Show your working clearly.​

Answer 1

Answer:

d = 2a

Step-by-step explanation:

      Sum of first 2n term is:

⇒ (2n/2) [2a + (2n - 1)d]

⇒ n[2a + (2n - 1)d]

       Sum of first n term is:

⇒ (n/2) [2a + (n - 1)d]

Given, S₂ₙ = 4 Sₙ

⇒ n[2a + (2n - 1)d] = 4[(n/2) [2a + (n - 1)d]

⇒ n[2a + (2n - 1)d] = 2n[2a + (n - 1)d]]

⇒ [2a + (2n - 1)d] = 2[2a + (n - 1)d])

⇒ 2a + 2nd - d = 4a + 2nd - 2d

⇒ 2a - d = 4a - 2d

⇒ 2d - d = 4a - 2a

⇒ d = 2a

Answer 2

Answer:

d =2a

Step-by-step explanation:

hope it is help you✍️✍️