Question

If the sum of the first $n^{th}$ 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

Step 1. Find the terms of an AP.

As we now that, an Arithmetic progression is a sequence of numbers such that the difference $d$ between each consecutive term is a constant.

The sequence is in the form: $\mathrm{a},\mathrm{a}+\mathrm{d},\mathrm{a}+2\mathrm{d},\mathrm{a}+3\mathrm{d},...$

We can find all given terms of an AP by using the following formula,

The $n^{th}$ term is given as,

${\mathbf{T}}_{\mathbf{n}}\mathbf{=}\mathbf{}\mathbf{a}\mathbf{+}\left(n-1\right)\mathbf{\times }\mathbf{d}$

Here, ${\mathrm{T}}_{\mathrm{n}}$ is the ${\mathrm{n}}^{\mathrm{th}}$ term, $\mathrm{a}$ is first term and $\mathrm{d}$ is the common difference.

The Sum of first ${\mathrm{n}}^{\mathrm{th}}$ terms,

${\mathbf{S}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{2}}\left[2a+\left(n-1\right)\times d\right]$

It is given that, the first $n^{th}$ terms of an AP is $4\mathrm{n}-{\mathrm{n}}^2$.

Let,

${\mathrm{S}}_{\mathrm{n}}=4\mathrm{n}-{\mathrm{n}}^2$ ----------- $\left(1\right)$

Put $n=1$ in equation $\left(1\right)$.

$\Rightarrow$ ${\mathrm{S}}_1=4\left(1\right)-{\left(1\right)}^2$

$\Rightarrow$ ${\mathrm{S}}_1=4-1$

$\Rightarrow$ ${\mathrm{S}}_1=3$

Therefore, the sum of first term of AP is $3$.

Step 2. Find the sum of first two terms.

Put $\mathrm{n}=2$ in equation $\left(1\right)$.

$\Rightarrow$ ${\mathrm{S}}_2=4\left(2\right)-{\left(2\right)}^2$

$\Rightarrow$ ${\mathrm{S}}_2=8-4$

$\Rightarrow$ ${\mathrm{S}}_2=4$

The sum of first two term is $4$.

But sum of first term will be the first term.

So, First term $\mathrm{a}=3$

Step 3. Calculate the second term.

From the above calculation, the sum of first two terms is $4$. So,

$\Rightarrow$ $\mathrm{First}\mathrm{term}+\mathrm{Second}\mathrm{term}=4$

$\Rightarrow$ $3+{\mathrm{a}}_2=4$

$\Rightarrow$ ${\mathrm{a}}_2=4-3$

$\Rightarrow$ ${\mathrm{a}}_2=1$

The common difference $\mathrm{d}$ is,

$\Rightarrow$ $\mathrm{d}=\mathrm{Second}\mathrm{term}-\mathrm{First}\mathrm{term}$

$\Rightarrow$ $\mathrm{d}=1-3$

$\Rightarrow$ $\mathrm{d}=-2$

Thus, the sequence of an AP is, $3,1,-1,-3,...$.

Step 4. Find the $3^{\mathrm{rd}}$ and ${10}^{\mathrm{th}}$ term.

As we know that, the formula for ${\mathrm{n}}^{\mathrm{th}}$ term is,

${\mathrm{T}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\times \mathrm{d}$

For third term,

$\Rightarrow$ ${\mathrm{T}}_3=\mathrm{a}+\left(3-1\right)\times \mathrm{d}$

$\Rightarrow$ ${\mathrm{T}}_3=\mathrm{a}+2\mathrm{d}$

$\Rightarrow$ ${\mathrm{T}}_3=3+2\times \left(-2\right)$

$\Rightarrow$ ${\mathrm{T}}_3=-1$

The third term of an AP is $-1$.

For tenth term,

$\Rightarrow$ ${\mathrm{T}}_{10}=\mathrm{a}+\left(10-1\right)\times \mathrm{d}$

$\Rightarrow$ ${\mathrm{T}}_{10}=\mathrm{a}+9\mathrm{d}$

$\Rightarrow$ ${\mathrm{T}}_{10}=3+9\times \left(-2\right)$

$\Rightarrow$ ${\mathrm{T}}_{10}=-15$

Step 5. Find the ${\mathrm{n}}^{\mathrm{th}}$ term of an AP.

The ${\mathrm{n}}^{\mathrm{th}}$term of an AP is,

$\Rightarrow$ ${\mathrm{T}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\times \mathrm{d}$

$\Rightarrow$ ${\mathrm{T}}_{\mathrm{n}}=3+\left(\mathrm{n}-1\right)\times \left(-2\right)$

$\Rightarrow$ ${\mathrm{T}}_{\mathrm{n}}=3-2\mathrm{n}+2$

$\Rightarrow$ ${\mathrm{T}}_{\mathrm{n}}=5-2\mathrm{n}$

Therefore, the sum of first two terms is $4$, the second term is $1$ and the $3^{\mathrm{rd}}$, the ${10}^{\mathrm{th}}$, and the ${\mathrm{n}}^{\mathrm{th}}$ terms are $-1,-15,$ and $5-2\mathrm{n}$ respectively.