Question

Please I need this before 12pm

Answer 1

Answer:

The required formula is:

                                                          ${\displaystyle \ a_{n}=a_{1}+(n-1)d}$

Step-by-step explanation:

The total number of squares of the the first term = 4

The total number of squares of the the second term = 7

The total number of squares of the the third term = 10

so,

$a_1=4$

$a_2=7$

$a_3=10$

Finding the common difference d

$d=a_3-a_2=10-7=3$

$d=a_2-a_1=7-4=3$

As the common difference 'd' is same, it means the sequence is in arithmetic.

So

If the initial term of an arithmetic progression is ${\displaystyle a_{1}}$ and the common difference of successive members is d, then the nth term of the sequence $(a_n)$ is given by:

                         ${\displaystyle \ a_{n}=a_{1}+(n-1)d}$

Therefore, the required formula is:

                                                          ${\displaystyle \ a_{n}=a_{1}+(n-1)d}$