Question

If a_n = 3 – 4n, show that a₁, a₂, a₃,... form an AP. Also find S₂₀.

Answer 1

Given that, n^th term of the series is

a_n = 3 – 4n   ...(i)

Put n = 1,

a₁ = 3 – 4(1)

= 3 – 4

= –1

Put n = 2,

a₂ = 3 – 4(2)

= 3 – 8

= –5

Put n = 3,

a₃ = 3 – 4(3)

= 3 – 12

= –9

Put n = 4,

a₄ = 3 – 4(4)

= 3 – 16

= –13

So, the series becomes –1, –5, –9, –13,...

We see that,

a₂ – a₁

= –5 – (–1)

= –5 + 1

= –4

a₃ – a₂

= –9 – (–5)

= –9 + 5

= –4

a₄ – a₃

= –13 – (–9)

= –13 + 9

= –4

i.e., a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = ... = –4

Since the each successive term of the series has the same difference.

So, it forms an AP.

We know that, sum of n terms of an AP,

S_n = `n/2[2a + (n - 1)d]`

∴ Sum of 20 terms of the AP,

S₂₀ = `20/2[2(-1) + (20 - 1)(-4)]`

= 10[–2 + (19)(–4)]

= 10(–2 – 76)

= 10 × (–78)

= –780

Hence, the required sum of 20 terms i.e., S₂₀ is –780.