Question

If the sum of the first n term of an AP is $4n-n^2$, what is the first term $\left(tahtis{S}_1\right)$? what is the sum of first two terms? what is the second term? Similar find the 3rd, the 10th and nth terms.

Answer 1

We have,

$S_n=4n-n^2$

Taking $n=1$ and we get,

$S_1=4\times 1-1^2$

$=4-1$

$S_1=3$

So, the sum of fristr term of an A.P. is $3$.

But, sum of first term will be the first term.

So, first term $a=3$.

Now, taking $n=2$ and we get,

$S_n=4n-n^2$

$S_2=4\times 2-2^2$

$S_2=4$

Then, sum of first $2$ terms is 4.

Now, finding $\text{Second}\text{term}$.

Then,

According to given question,

$\text{First}\text{term}\text{+Second}\text{term}=4$

$3+a_2=4$

$a_2=1$

Hence the $A.P.is$

First term $a=3$

Common difference

$d=\text{Second}\text{term}\text{-first}\text{term}$

$\text{d=1-3}$

$\text{=-2}$

The third term of

$a_3=a+2d$

$a_3=3+2\times (-2)$

$a_3=-1$

Similarly, ${10}^{th}term$

$a_{10}=a+9d$

$=3+9(-2)$

$=3-18$

$=-15$

Then,

$n^{th}termis$

$a_n=a+(n-1)d$

$a_n=3+(n-1)(-2)$

$a_n=3-2n+2$

$a_n=5-2n$

Hence, this is the answer.