Question

An arithmetic series consists of consecutive integers that are multiples are of 4. What is the sum of the first nine terms of this sequence if the first term is 0
A. 160


B. 162


C. 180


D. 198

Answer 1

Answer:

C. 180

Step-by-step explanation:

We are given that the arithmetic series consists of integers that are multiples of 4 with first term 0.

So,we get the series as,

0, 4, 8, 12, 16, 20, 24, 28, 32, 36.

Now, it is required to find the sum of first nine non-zero terms.

We have the sum of the arithmetic series is given by,

$S_{n}=\frac{n}{2} \times [2a+(n-1)d]$

where a = 4 and d = 8-4 = 4.

Substituting the values, we get,

$S_{n}=\frac{n}{2} \times [2 \times 4+4(n-1)]$

As, we want the sum of first nine non-zero terms, this gives n=9

So, $S_{9}=\frac{9}{2} \times [2 \times 4+4(9-1)]$

i.e. $S_{9}=\frac{9}{2} \times [2 \times 4+4 \times 8]$

i.e. $S_{9}=\frac{9}{2} \times [8+32]$

i.e. $S_{9}=\frac{9}{2} \times [40]$

i.e. $S_{9}=9 \times 20$

i.e. $S_{9}=180$

Hence, the sum of first nine non-zero terms is 180.

Answer 2

Answer:  The answer is 144.

Step-by-step explanation: Let 'a' be the first term and 'd' be the common difference of the given arithmetic progression (A.P.).

According to the given information, A.P. is {0, 4, 8, 12, 16, . . .}, i.e., a=0 and d=4.

The sum of first n terms is given by

$S_n=\dfrac{n}{2}\{2a+(n-1)d\}.$

So, the sum of first 9 terms is

$S_9=\dfrac{9}{2}\{2\times 0+(9-1)\times4\}=\dfrac{9}{2}(8\times 4)=144.$

Thus, the required sum is 144.