If the sum of first n terms of an AP is 4n – n2, what is the first term i.e. S1? What is the sum of the first two terms? What is the second term? Similarly, find the 3rd, then 10th, and nth terms.

Table of Contents

Solution:

 

Given, Sn = 4n – n2

Therefore, S1 = 4(1) – (1)2 = 4 – 1 = 3 (Answer)

And S2 = 4(2) – (2)2 = 8 – 4 = 4

∴ Second term = a2

= S2 – S1 [∵ an = Sn – Sn – 1]

= 4 – 3

= 1 (Answer)

Similarly, S3 = 4(3) – (3)2 = 12 – 9 = 3(Answer)

S9 = 4(9) – (9)2 = 36 – 81 = -45

S10 = 4(10) – (10)2 = 40 – 100 = -60

Therefore,

Third term = a3 = S3 – S2 [∵ an = Sn – Sn – 1]

= 3 – 4

= -1 (Answer)

Tenth term = a10 = S10 – S9 [∵ an = Sn – Sn – 1]

= -60 – (-45)

= -60 + 45

= -15 (Answer)

nth term = an = Sn – Sn – 1

= 4n – n2 – {4(n – 1) – (n – 1)2}

= 4n – n2 – {4n – 4 – (n2 – 2n + 1)}

= 4n – n2 – {4n – 4 – n2 + 2n – 1}

= 4n – n2 – (- n2 + 6n – 5)

= 4n – n2 + n2 – 6n + 5

= 5 – 2n (Answer)

Previous Show that a1, a2,…, an form an AP where an is defined as below.an = 9 – 5n. Also, find the sum of the first 15 terms in each case.

Leave a Reply