Question

If the sum of the first n terms of an AP is $4n – n^2$, what is the first term (that is $S_1$)? What is the sum of first two terms? What is the second term? Similarly, find the $3^{rd}$, the $10^{th}$ and the $n^{th}$ terms.

Answer 1

Given that,
$S_n = 4n − n^2$
First term, $a_1 = S_1 = 4(1) − (1)^2 = 4 − 1 = 3$
Sum of first two terms, $S_2 = 4(2) − (2)^2 = 8 − 4 = 4$
Second term, $a_2 = S_2 − S_1 = 4 − 3 =$1
$d = a_2 − a_1 = 1 − 3 = −2$
$a_n = a + (n − 1)d$
$a_n = 3 + (n − 1) (−2)$
$a_n = 3 − 2n + 2$
$a_n = 5 − 2n$
Therefore,
$a_3 = 5 − 2(3) = 5 − 6 = −1$
$a_{10} = 5 − 2(10) = 5 − 20 = −15$

Hence, the sum of first two terms is 4. The second term is 1. 3rd, 10th, and nth terms are −1, −15, and 5−2n respectively.