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

It is given that the sum of n terms of an AP is equal to (4n− $n^2$).

It means $S_n$=4n− $n^2$

Let’s calculate $S_1$ and $S_2$ using $S_n$=4n− $n^2$.

$S_1$=4(1) − (1)2=4−1=3

$S_2$=4(2) − (2)2=8−4=4

First term =a=$S_1$=3 (1)

Let’s find common difference now.

We can write any AP in the form of general terms like a, a +d, a + 2d...

We have calculated that sum of first two terms is equal to 4 i.e. $S_2$=4

Therefore, we can say that a+ (a+ d) =4

Putting value of a from equation (1), we get

2a+d=4

$\Rightarrow$2 (3) +d=4

$\Rightarrow$6+d=4

$\Rightarrow$d=−2

Using formula $a_n$=a+ (n−1) d, to find nth term of AP, we can say that

Second term of AP = $a_2$ = a+ (2−1) d =3+ (2−1) (−2) =3−2 =1

Third term of AP = $a_3$ = a+ (3−1) d =3+ 2 (−2) =3−4=−1

Tenth term of AP = $a_{10}$ = a+ (10−1) d=a+ 9d =3+9 (−2) =3−18=−15

$n^{th}$ term of AP = $a_n$ = a+ (n−1) d

=3+ (n−1) (−2)

=3−2n+2

=5−2n