Question

Show that a₁, a_2,…,a_n, …. form an AP where a_n is defined as below: 

i) a = 3 + 4n 

ii) a_n = 9 – 5n. Also find the sum of the first 15 terms in each case.

Answer 1

i) a_n = 3 + 4n

Given: a_n = 3 + 4n 

Then a₁ = 3 + 4 × l = 3 + 4 = 7 

a₂ = 3 + 4 × 2 = 3 + 8 = 11 

a₃ = 3 + 4 × 3 = 3 + 12 = 15 

a₄ = 3 + 4 × 4 = 3 + 16 = 19 

Now the pattern is 7, 11, 15, …… 

where a = a₁ = 7; a₂ = 11; a₃ = 15, ….. and 

a₂ – a₁ = 11 – 7 = 4; 

a₃ – a₂ = 15 – 11 = 4; 

Here d = 4 Hence a₁, a₂, ….., a_n….. forms an A.P.

ii) a_n = 9 – 5n 

Given: a_n = 9 – 5n. 

a₁ = 9 – 5 × l = 9 – 5 = 4 

a₂ = 9 – 5 × 2 = 9 – 10 = -1 

a₃ = 9 – 5 × 3 = 9 – 15 = -6

a₄ = 9 – 5 × 4 = 9 – 20 = -11 

Also 

a₂ – a₁ = -1 – 4 = -5; 

a₃ – a₂ = -6 – (-1) = – 6 + 1 = -5 

a₄ – a₃ = -11 – (-6) = -11 + 6 = -5 

∴ d = a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = …. = -5

Thus the difference between any two successive terms is constant (or) starting from the second term, each term is obtained by adding a fixed number ‘-5’ to its preceding term. Hence {a_n} forms an A.P.