Question

Find the sum of $16.2, 5.4, 1.8,...$ upto 7 terms.

Answer 1

A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed ,non-zero number called the common ratio.
Sum to n terms of the geometric sequence or progression is given as $= {S_n} = a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right)$
Where $a$=first term of the sequence
$r$= common difference between the consecutive terms.

Complete step by step solution:
Clearly the given series is in geometric progression where the common ratio is $= \dfrac{{5.4}}{{16.2}} = \dfrac{1}{3}$.
Using formula:
$= {S_n} = a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right)$
Where $a$=first term of the sequence
$r$= common difference between the consecutive terms.
$\Rightarrow {S_7} = 16.2 \times \dfrac{{2186}}{{2187}}$
$n$= the number of terms upto which the summation of series is needed.
${S_n}$=Summation of series upto n terms.

$$\Rightarrow {S_7} = 16.2\dfrac{{\left( {1 - {{\left( {\dfrac{1}{3}} \right)}^7}} \right)}}{{\left( {1 - \dfrac{1}{3}} \right)}} \\\ \Rightarrow {S_7} = 16.2\dfrac{{\left( {1 - \dfrac{1}{{2187}}} \right)}}{{\left( {\dfrac{2}{3}} \right)}} \\\ \Rightarrow {S_7} = 16.2 \times \dfrac{{2186}}{{2187}} \\\ \Rightarrow {S_7} = \dfrac{{1093}}{{45}} \\\$$

The sum of $16.2, 5.4, 1.8,...$upto 7 terms is $\dfrac{{1093}}{{45}}$.

Note:
In such questions first look for the sequence whether it is an arithmetic sequence or geometric progression. Then go through the question and apply the most appropriate formula.
In maximum cases, if the sequence is increasing or decreasing at a higher rate it will be a geometric progression or sequence.