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

Question

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

Becky Harrison · Accepted Answer

Given data is  a1=18  common ratio (r)=2  We have to calculate 4-th partial sum of CnP.  S=a1(1−rn1−r)  By putting the values in formulae.  ⇒S=18(1−241−2)  ⇒S=18(1−161−2)  ⇒S=18(−15−1)  ⇒S=18×15=270  So the answer is 270.

deginasiba · Accepted Answer

We have, sum to n terms of a geometric sequence is given by.  Sn=a1{(r)n−1}r−1 for r&amp;gt;1  So, putting given values n=4, we get  S4=18[(2)4−1]2−1=18(16−1)1=18×15=270  Hence, 4th partial sum =S4=270  Option c is correct

user_27qwe · Accepted Answer

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

What is the 4th partial sum for the geometric sequence

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