
The sum of a few terms of any ratio series is 728, if the common ratio is 3 and the last term is 486, then the first term of the series will be
(1) 2
(2) 1
(3) 3
(4) 4
Answer
233.1k+ views
Hint: A sort of sequence known as geometric progression (GP) is one in which each following phrase is created by multiplying each preceding term by a fixed number, or "common ratio." This progression is sometimes referred to as a pattern-following geometric sequence of numbers.
Formula Used:
Common ratio is $={{a}_{n}}/{{a}_{n-1}}$
Sum to n terms of GP $S_{n}=a\left(\dfrac{r^{n}-1}{r-1}\right)$
Complete step by step Solution:
The common ratio in the geometric progression is the ratio obtained by dividing each phrase in the series by the term before it. Typically, the letter "r" is used to indicate it.
The geometric progression's common ratio calculation formula is: $a, a r, a r^{2}, a r^{3}$
Common ratio is $=a_{n} / a_{n-1}$
Here, common ratio, $r=3$
$n^{\text {th }} \text { term, } a_{n}=486$
$S_{n}=728$
$a_{n}=486$
$\Rightarrow a r^{n-1}=486$
Substitute the value $r=3$
$\Rightarrow a(3)^{n-1}=486$
$\Rightarrow a(3)^{n}=486 \times 3$
$\Rightarrow a{{(3)}^{n}}=1458....\ldots (i)$
We know that sum to n terms of the series is
$S_{n}=728$
Then substitute the values of $S_{n}=728$ and $r=3$ in the formula
$\Rightarrow 728=a\left(\dfrac{3^{n}-1}{3-1}\right)$
$\Rightarrow 728=\left\{\dfrac{a(3)^{n}-a}{2}\right\}$
Simplify the expression
$\Rightarrow 1456=a(3)^{n-1}-a$
$\Rightarrow 1456=1458-a[$ From $(i)]$
$\Rightarrow a=1458-1456$
$\Rightarrow a=2$
The first term of the series is $2$
Hence, the correct option is 1.
Note: When a geometric progression (GP) is finite, it may be represented as $a,a{{r}^{2}},a{{r}^{3}},\ldots a{{r}^{n-1}}$and when it is infinite, it can be expressed as$a,a{{r}^{2}},a{{r}^{3}},\ldots a{{r}^{n-1}}$. Using a few formulas, we can determine the sum to n terms of GP for both finite and infinite GP. Additionally, the equation for calculating the sum of the finite and finite GP separately can be derived. The first n terms of a GP are collectively referred to as the total to n terms of a GP.
Formula Used:
Common ratio is $={{a}_{n}}/{{a}_{n-1}}$
Sum to n terms of GP $S_{n}=a\left(\dfrac{r^{n}-1}{r-1}\right)$
Complete step by step Solution:
The common ratio in the geometric progression is the ratio obtained by dividing each phrase in the series by the term before it. Typically, the letter "r" is used to indicate it.
The geometric progression's common ratio calculation formula is: $a, a r, a r^{2}, a r^{3}$
Common ratio is $=a_{n} / a_{n-1}$
Here, common ratio, $r=3$
$n^{\text {th }} \text { term, } a_{n}=486$
$S_{n}=728$
$a_{n}=486$
$\Rightarrow a r^{n-1}=486$
Substitute the value $r=3$
$\Rightarrow a(3)^{n-1}=486$
$\Rightarrow a(3)^{n}=486 \times 3$
$\Rightarrow a{{(3)}^{n}}=1458....\ldots (i)$
We know that sum to n terms of the series is
$S_{n}=728$
Then substitute the values of $S_{n}=728$ and $r=3$ in the formula
$\Rightarrow 728=a\left(\dfrac{3^{n}-1}{3-1}\right)$
$\Rightarrow 728=\left\{\dfrac{a(3)^{n}-a}{2}\right\}$
Simplify the expression
$\Rightarrow 1456=a(3)^{n-1}-a$
$\Rightarrow 1456=1458-a[$ From $(i)]$
$\Rightarrow a=1458-1456$
$\Rightarrow a=2$
The first term of the series is $2$
Hence, the correct option is 1.
Note: When a geometric progression (GP) is finite, it may be represented as $a,a{{r}^{2}},a{{r}^{3}},\ldots a{{r}^{n-1}}$and when it is infinite, it can be expressed as$a,a{{r}^{2}},a{{r}^{3}},\ldots a{{r}^{n-1}}$. Using a few formulas, we can determine the sum to n terms of GP for both finite and infinite GP. Additionally, the equation for calculating the sum of the finite and finite GP separately can be derived. The first n terms of a GP are collectively referred to as the total to n terms of a GP.
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