Question

Find the sum of first $51$ terms of an AP whose second and third terms are $14$ and $18$ respectively.

Answer 1

Find the sum of all the terms of the series.

Second term, $a_2=14$

Third term, $a_3=18$

We need to find the sum of first $51$ terms of the given series.

Common difference, $d=a_3-a_2=18-14=4$

$$\begin{gathered} \Rightarrow a_2=a+d \\ \Rightarrow 14=a+4 \end{gathered}$$

$a=10=$ First term

Sum of$n$ terms;

$$\begin{gathered} \Rightarrow S_n=\frac{n}{2}[2a+(n–1\left)d\right] \\ \Rightarrow S_{51}=\frac{51}{2}[2\times 10+(51-1)4] \\ \Rightarrow S_{51}=\frac{51}{2}[20+(50)\times 4] \\ \Rightarrow S_{51}=\frac{51}{2}\times [20+200] \\ \Rightarrow S_{51}=51\times 110 \\ \Rightarrow S_{51}=5610 \end{gathered}$$

Hence, the sum of first$51$ terms is $5610$.