Question

Finda) The sum of the first 10 termsb) the first term and the common difference

Answer 1

EXPLANATION

As the sum is given by the arithmetic sequence:

$S_n=\frac{3n(n-33)}{2}$

a)

Applying the sumatory to the first 10 terms:

$S_{10}=\frac{3\cdot10(10-33)}{2}$

Subtracting numbers:

$S_{10}=\frac{3\cdot10\cdot(-23)}{2}$

Multiplying numbers:

$S_{10}=\frac{-690}{2}$

Simplifying:

$-345$

b) The first term is given as shown as follows:

$S_1=\frac{3\cdot1\cdot(1-33)}{2}=\frac{-96}{2}=-48$

We can get the common difference by computing each subsequent number of the sequence and subtracting to the last one as shown as follows:

$a_2=S_2-(-48)=\frac{3\cdot2\cdot(2-33)}{2}-(-48)=-93+48=-45$

Now, subtracting -45 to the first term -48 give us:

-45 - (-48) = -45 + 48 = -3

In conclusion, the common difference is -3