In an A.P. the su

Let the first term of AP be ‘a’ and common difference be ‘d’.

We know, sum of ‘n’ terms of an AP is

Given,

Sum of first test ten terms, S₁₀ = 80 ……[1]

⇒ 5(2a + 9d) = 80

⇒ 2a + 9d = 16 ……[2]

Also, sum of next ten terms = 280

⇒ S₂₀ – S₁₀ = 280

⇒ S₂₀ – 80 = 280 [From 1]

⇒ S₂₀ = 360

⇒ 2a + 19d = 36 ……[3]

On subtracting [2] from [3]

2a + 19d – 2a – 9d = 36 – 16

⇒ 10d = 20

⇒ d = 2

Putting d = 2 in [2]

⇒ 2a + 18 = 16

⇒ a = –1

Hence, AP is

–1, –1 + 2, –1 + 2(2), …

–1, 1, 3, …