# If an = 3 – 4n, show that 123 a1, a2, a3,... form an AP. Also find S20 - Mathematics

Sum

If an = 3 – 4n, show that 123 a1, a2, a3,... form an AP.

Also find S20

#### Solution

Given that, nth term of the series is

an = 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

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., S20 is – 780

Concept: Sum of First n Terms of an A.P.
Is there an error in this question or solution?

#### APPEARS IN

NCERT Mathematics Exemplar Class 10
Chapter 5 Arithematic Progressions
Exercise 5.3 | Q 23 | Page 53
Share