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Find the sum of the first 48 terms of the following series, to the nearest integer.

6, 9,12,...
6,9,12

User Leslie Wu
by
5.9k points

1 Answer

7 votes

Answer:

3,672

Explanation:

Given the sequence 6, 9, 12...

The sequence is an arithmetic sequence

first term a = 6

common difference d = 9 - 6 = 12 - 9 = 3

number of terms n = 48

Sn = n/2[2a+(n-1)d]

Substitute the given values

S48 = 48/2[2(6)+(48-1)(3)]

S48 = 24(12+(3*47))

S48 = 24(12+141)

S48 = 24(153)

S48 = 3,672

Hence the sum of the first 48terms is 3,672

User Alphonse Prakash
by
6.0k points