Verify that a, 2a + 1, 3a + 2, 4a + 3,…is an AP, and then write i

1.	Verify that a, 2a + 1, 3a + 2, 4a + 3,…is an AP, and then write its next three terms.
Answer» Here a₁ = a a₂ = 2a + 1 a₃ = 3a + 2 a₄ = 4a + 3 a₂ – a₁ = (2a + 1) – (a) = a + 1 a₃ – a₂ = (3a + 2) – (2a + 1) = a + 1 a₄ – a₃ = (4a + 3) – (3a+2) = a + 1 Since, difference of successive terms are equal, Hence, a, 2a + 1, 3a + 2, 4a + 3,… is an AP with common difference a+1. Therefore, the next three term will be, 4a + 3 +(a + 1), 4a + 3 + 2(a + 1), 4a + 3 + 3(a + 1) 5a + 4, 6a + 5, 7a + 6

Verify that a, 2a + 1, 3a + 2, 4a + 3,…is an AP, and then write its next three terms.

Answer»

Here

a₁ = a

a₂ = 2a + 1

a₃ = 3a + 2

a₄ = 4a + 3

a₂ – a₁ = (2a + 1) – (a) = a + 1

a₃ – a₂ = (3a + 2) – (2a + 1) = a + 1

a₄ – a₃ = (4a + 3) – (3a+2) = a + 1

Since, difference of successive terms are equal,

Hence, a, 2a + 1, 3a + 2, 4a + 3,… is an AP with common difference a+1.

Therefore, the next three term will be,

4a + 3 +(a + 1), 4a + 3 + 2(a + 1), 4a + 3 + 3(a + 1)

5a + 4, 6a + 5, 7a + 6