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Question: How do you find the \( 10th \) term in the geometric sequence \( 320,160,80,40,... \) ?...

How do you find the 10th10th term in the geometric sequence 320,160,80,40,...320,160,80,40,... ?

Explanation

Solution

We know that Geometric Progression is a type of sequence where each succeeding term is produced by multiplying each consecutive term by a fixed number, which is known as a common ratio. This progression is also known as a geometric sequence of numbers that follow a pattern. Therefore, we will first find the common ratio and then apply the formula for finding the nthnth term of the geometric sequence.

Formulas used: r=a2a1r = \dfrac{{{a_2}}}{{{a_1}}} , where, rr is the common ratio, a2{a_2} is the second term of the geometric sequence and a1{a_1} is the first term of the geometric sequence
an=a1rn1{a_n} = {a_1}{r^{n - 1}} , where, an{a_n} is the nthnth term of the geometric sequence, a1{a_1} is the first term of the geometric sequence and rr is the common ratio

Complete step-by-step solution:
Here, we are given the geometric series 320,160,80,40,...320,160,80,40,...
We have the first term a1=320{a_1} = 320 and the second term a2=160{a_2} = 160 .
Now, we will find the common ratio.
r=a2a1=160320=12r = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{160}}{{320}} = \dfrac{1}{2}
We know that the formula for finding the nthnth term of the geometric sequence is an=a1rn1{a_n} = {a_1}{r^{n - 1}} .
Therefore, for finding the 10th10th term, we will use the formula
a10=320×(12)101=320×(12)9=320×(129){a_{10}} = 320 \times {\left( {\dfrac{1}{2}} \right)^{10 - 1}} = 320 \times {\left( {\dfrac{1}{2}} \right)^9} = 320 \times \left( {\dfrac{1}{{{2^9}}}} \right)
We can write 320=10×32=10×25320 = 10 \times 32 = 10 \times {2^5}
a10=10×25×(129)=10×1295=10×124=1016=58{a_{10}} = 10 \times {2^5} \times \left( {\dfrac{1}{{{2^9}}}} \right) = 10 \times \dfrac{1}{{{2^{9 - 5}}}} = 10 \times \dfrac{1}{{{2^4}}} = \dfrac{{10}}{{16}} = \dfrac{5}{8}
Thus, the 10th10th term in the geometric sequence 320,160,80,40,...320,160,80,40,... is 58\dfrac{5}{8} .

Note: We have seen that a geometric progression or a geometric sequence in which each term is varied by another by a common ratio. The next term of the sequence is produced when we multiply a constant which is a non-zero value, to the preceding term. In other words, we can also say that when we divide any succeeding term from its preceding term, then we get the value equal to common ratio.