Anasthaesium wrote:
How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) +.... will form 120 + 121*(3^(1/2))?
A. 3
B. 4
C. 5
D. 6
E. 9
We have a following sequence:
\sqrt{3},
3,
3\sqrt{3},
9, ...
Notice that this sequence is a geometric progression with first term equal to
\sqrt{3} and common ratio also equal to
\sqrt{3}.
The question asks: the sum of how many terms of this sequence adds up to
120+121\sqrt{3}.
The sum of the first
n terms of geometric progression is given by:
sum=\frac{b*(r^{n}-1)}{r-1}, (where
b is the first term,
n # of terms and
r is a common ratio
\neq{1}).
So,
\frac{\sqrt{3}(\sqrt{3}^n-1)}{\sqrt{3}-1}=120+121\sqrt{3} -->
\sqrt{3}(\sqrt{3}^n-1)=120\sqrt{3}+121*3-120-121\sqrt{3} -->
\sqrt{3}*\sqrt{3}^n-\sqrt{3}=243-\sqrt{3} -->
\sqrt{3}*\sqrt{3}^n=243 -->
3^{\frac{1+n}{2}}=3^5 -->
\frac{1+n}{2}=5 -->
n=9.
Answer: E.
Shortcut solution:Alternately you can spot that
every second term (which also form a geometric progression), 3, 9, 27, ... should add up to 120:
\frac{3(3^k-1)}{3-1}=120 -->
3^k-1=80 -->
3^k=81 -->
k=4, so the number of terms must be either
2k=8 (if there are equal number of irrational and rational terms) or
2k+1=9 (if # of irrational terms, with
\sqrt{3}, is one more than # rational terms), since only 9 is present among options then it must be a correct answers.
Hope it's clear.
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56% (03:06) correct
