Question

If the sum of first $7$ terms of an AP is $49$ and that of $17$ terms is $289$, find the sum of first $n$ terms.

Answer 1

Step 1: Find the equation for first $7$ term

Given that, sum of the first $7$ terms is $49$

We know sum of $n$ terms;

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

Where $a$ is the first term and $d$ is the common difference

$$\begin{gathered} \therefore {\mathrm{S}}_7=\frac{7}{2}\left[2\mathrm{a}+\left(7-1\right)\mathrm{d}\right] \\ \Rightarrow 49=\frac{7}{2}\left[2\mathrm{a}+6\mathrm{d}\right] \\ \Rightarrow 49=\frac{7}{2}\times 2\left[\mathrm{a}+3\mathrm{d}\right] \\ \Rightarrow 49=7\left[\mathrm{a}+3\mathrm{d}\right] \\ \Rightarrow 7=a+3d....\left(i\right) \end{gathered}$$

Step 2: Find the equation for first $17$ term

Given that, sum of the first $17$ terms is $289$

$$\begin{gathered} \therefore {\mathrm{S}}_{17}=\frac{17}{2}\left[2\mathrm{a}+\left(17-1\right)\mathrm{d}\right] \\ \Rightarrow 289=\frac{17}{2}\left[2\mathrm{a}+16\mathrm{d}\right] \\ \Rightarrow 289=\frac{17}{2}\times 2\left[\mathrm{a}+8\mathrm{d}\right] \\ \Rightarrow 289=17\left[\mathrm{a}+8\mathrm{d}\right] \\ \Rightarrow 17=a+8d.....\left(ii\right) \end{gathered}$$

Step 3: Solving the equation

Subtracting equation $\left(ii\right)-\left(i\right)$

$$\begin{gathered} \Rightarrow \mathrm{a}+8\mathrm{d}-(\mathrm{a}+3\mathrm{d})=17-7 \\ \Rightarrow \mathrm{a}+8\mathrm{d}-\mathrm{a}-3\mathrm{d}=10 \\ \Rightarrow 5\mathrm{d}=10 \\ \Rightarrow \mathrm{d}=2 \end{gathered}$$

Put the value of d in equation $\left(i\right),$

$$\begin{gathered} \Rightarrow \mathrm{a}+3\left(2\right)=7 \\ \Rightarrow \mathrm{a}+6=7 \\ \Rightarrow \mathrm{a}=7-6 \\ \Rightarrow \mathrm{a}=1 \end{gathered}$$

Step 4: Find the sum of first $n$ terms

Therefore $a=1,d=2$

We know that the sum of first $n$ term of an A.P series is given by ${\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\right]$

$$\begin{gathered} \therefore {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\right] \\ \Rightarrow {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\left(1\right)+(\mathrm{n}-1)2\right] \\ \Rightarrow {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2+(\mathrm{n}-1)2\right] \\ \Rightarrow {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2+2\mathrm{n}-2\right] \\ \Rightarrow {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{n}\right] \\ \Rightarrow {\mathrm{S}}_{\mathrm{n}}={\mathrm{n}}^2 \end{gathered}$$

Hence, Sum of n terms is $n^2$