If a_{n} = 3 – 4n, show that 123 a_{1}, a_{2}, a_{3},... form an AP.

Also find S_{20}

#### Solution

Given that, n^{th} term of the series is

a_{n} = 3 – 4n .......(i)

Put n = 1, a_{1} = 3 – 4(1) = 3 – 4 = –1

Put n = 2, a_{2} = 3 – 4(2) = 3 – 8 = –5

Put n = 3, a_{3} = 3 – 4(3) = 3 – 12 = –9

Put n = 4, a_{4} = 3 – 4(4) = 3 – 16 = –13

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

We see that,

a_{2} – a_{1} = –5 – (–1) = –5 + 1 = – 4

a_{3} – a_{2} = –9 – (–5) = –9 + 5 = – 4

a_{4} – a_{3} = -13 – (–9) = –13 + 9 = – 4

i.e., a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{3} = … = – 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

`S_20 = 20/2[2(-1) + (20 - 1)(-4)]`

= `10[-2 + (19)(-4)]`

= `10(-2 - 76)`

= `10 xx (-78)`

= `- 780`

Hence, the required sum of 20 terms i.e., S_{20} is – 780