Question

Find the sum of the arithmetic progression 1,4,7,10,13,16 .Every third term of the above progression is removed i.e 7,16. Find the sum of the remaining terms​

Answer 1

Answer:

The sum of the arithmetic progression is 51

The sum of the remaining terms is 28

Step-by-step explanation:

Sum of terms of an arithmetic progression

Being a1 the first term of an arithmetic progression, an its last term, and n the number of terms, the sum of all terms is calculated as:

$\displaystyle S_n=\frac{a_1+a_n}{2}\cdot n$

We are given the progression:

1,4,7,10,13,16

We need to check if the list of numbers is a progression. The difference between consecutive terms must be equivalent.

Since:

4 - 1 = 3

7 - 4 = 3

10 - 7 = 3

And the rest of the differences also result in 3, the series of numbers has a common difference of 3.

The progression has the following parameters:

a1=1, an=16, n=6

Let's find the sum:

$\displaystyle S_6=\frac{1+16}{2}\cdot 6$

$\displaystyle S_6=\frac{17}{2}\cdot 6$

$S_6=51$

The sum of the arithmetic progression is 51

If we removed the terms 7 and 16, the new sum is:

51 - 7 - 16 = 28

The sum of the remaining terms is 28