Question

Find the sum of the first 47 terms of the following series, to the nearest integer.

Answer 1

Answer:

The sum of the first 47 terms of the series, 12, 16, 20, ... S₄₈ is 4,888

Step-by-step explanation:

The given series is;

12, 16, 20, ...,

Therefore, the first term of the series is, a = 12

The common difference of series is found as follows;

The difference between subsequent terms, 12 and 16 is  16 - 12 = 4

The difference between subsequent terms, 16, and 20 is  20 - 16 = 4

Therefore, the common difference, d = 4

The series is therefore an arithmetic projection, AP

The sum of the first 'n' terms of an AP, Sₙ, is given as follows;

$S_n = \dfrac{n}{2} \cdot \left [2 \cdot a + (n - 1)\cdot d \right ]$

(47/2)*(2*12+(47-1)*4)

The sum of the first 47 terms is therefore given as follows;

$S_n = \dfrac{47}{2} \cdot \left [2 \times 12 + (47 - 1)\times 4 \right ] = 4,888$

The sum of the first 47 terms of the series, 12, 16, 20, ... S₄₈ = 4,888