What is the 4th partial sum for the geometric sequence

Irvin Dukes

Irvin Dukes

Answered question

2021-12-29

What is the 4th partial sum for the geometric sequence where a1=18 and r=2
a) S4=18
b) S4=288
c) S4=152
d) S4=270

Answer & Explanation

Becky Harrison

Becky Harrison

Beginner2021-12-30Added 40 answers

Given data is
a1=18
common ratio (r)=2
We have to calculate 4-th partial sum of CnP.
S=a1(1rn1r)
By putting the values in formulae.
S=18(12412)
S=18(11612)
S=18(151)
S=18×15=270
So the answer is 270.
deginasiba

deginasiba

Beginner2021-12-31Added 31 answers

We have, sum to n terms of a geometric sequence is given by.
Sn=a1{(r)n1}r1 for r>1
So, putting given values n=4, we get
S4=18[(2)41]21=18(161)1=18×15=270
Hence, 4th partial sum =S4=270
Option c is correct
user_27qwe

user_27qwe

Skilled2022-01-05Added 375 answers

The 4th partial sum of the sequence is the sum of the first 4 terms:
S4=a(1)+a(2)+a(3)+a(4)
Because the sequence is geometric, any term can be written in terms of the previous one:
a(n) = r a(n - 1) and, more importanly, in terms of the first term a(1), so that
a(2)=ra(1)
a(3)=ra(2)=r2a(1)
a(4)=ra(3)=r3a(1)
Then the 4th partial sum is
S4=a(1)+ra(1)+r2a(1)+r3a(1)
S4=a(1)(1+r+r2+r3)
With a(1) = 18 and r = 2 we have
S4=18(1+2+4+8)=270

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