Question

Find the sum of first 51 terms of an AP whose 2nd and 3rd terms are 14 and 18 respectively​

Answer 1

Answer:

The sum of the first 51 terms of the AP is 5610.

Step-by-step explanation:

From the given, the second term is 14 and the third term is 18.

We can call the first term of the AP as 'a'. And the common difference is called 'd'. To find the common difference subtract the second term from the third term. That is,

                    d = 18 - 14  

                    d = 4

To find the first term subtract 'd' from the second term. Therefore, a = 10

The formula to find the sum of n terms is given as,

                           $S_{n}=\dfrac{n}{2} (2a+\left( n-1\right) d)$

 The question is to find 51 terms, that is n = 51. Substitute the values in the equation.

hence,  $S_{n}=\dfrac{51}{2} (2\times 10+\left( 51-1\right) 4)$

            $\begin{array}{l}S_n=\dfrac{51}{2}\left(20+50\times4\right)\\\\\ \ \ \ \ \begin{aligned}=\dfrac{51}{2}\left(20+200\right)\\\\=\dfrac{51}{2}\left( 220\right) \end{aligned}\end{array}$

                      $=51\times 110$

                   $S_{n}=5610$

Therefore the sum of the first 51 terms of the AP is 5610.