Question

the first term of the G.P. is 50 and the fourth term is 1350. Find its 5th term

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{First term of a G.P is 50 and the fourth term is 1350}$

$\underline{\textbf{To find:}}$

$\textsf{5 th term of the G.P}$

$\underline{\textbf{Solution:}}$

$\begin{array}{c||c}\mathsf{t_1=50}&\mathsf{t_4=1350}\\\\\mathsf{a=50}&\mathsf{a\,r^3=1350}\\\\&\mathsf{(50)\,r^3=1350}\\\\&\mathsf{r^3=\dfrac{1350}{50}}\\\\&\mathsf{r^3=27}\\\\&\mathsf{r^3=3^3}\\\\&\boxed{\mathsf{r=3}}\end{array}$

$\mathsf{Now,}$

$\mathsf{t_5=ar^4}$

$\mathsf{t_5=50(3)^4}$

$\mathsf{t_5=50(81)}$

$\boxed{\bf\,t_5=4050}}$

$\underline{\textbf{Answer:}}$

$\textbf{5 th term is 4050 }$

$\underline{\textbf{Formula used:}}$

$\boxed{\begin{minipage}{6cm} \\\mathsf{The\;n\,th\;term\;of\;G.P,\;a,ar,ar^2\;.\;.\;.\;.\;\;is}\\\\\mathsf{\;\;\;\;t_n=a\,r^{n-1}}\\ \end{minipage}}$

Answer 2

Its 5th term is, 4050.

What is Geometric Progression?

• Geometric Progression (GP) is a type of sequence where each succeeding term is produced by multiplying each preceding term by a fixed number, which is called a common ratio.
• This progression is also known as a geometric sequence of numbers that follow a pattern.

How do you find the nth term of a GP?

• The nth term of a GP series is Tn = arn-1, where a = first term and r = common ratio = Tn/Tn-1) .
• The sum of infinite terms of a GP series S∞= a/(1-r) where 0< r<1. If a is the first term, r is the common ratio of a finite G.P.
• consisting of m terms, then the nth term from the end will be = arm-n.

According to the question:

Given:

First-term of a GP =50

The fourth term of a GP=1350

Formula used:

The $\mathrm{n}^{\text {th }} term of a \mathrm{GP}=a r^{\mathrm{n}-1}$

Where, $\mathrm{GP}= geometric progression, r= common ratio, n=n^{\text {th }} term$

Calculation:

a=50, and let r be the common ratio, $4^{\text {th }}$term =1350

$\begin{aligned}&\Rightarrow a_{4}=a r^{n-1} \\&\Rightarrow 1350=50 \times r^{(4-1)} \\&\Rightarrow r^{3}=27 \\&\Rightarrow r=3 \\&a_{5}=a r^{n-1}\end{aligned}$

$\Rightarrow a_{5}=50 \times 3^{(5-1)}=50 \times 3^{4} \Rightarrow a_{5}=50 \times 81=4050 \\ \\\therefore 5^{\text {th }} of a GP is 4050$

Hence, 4050 is its fifth term.

Learn more about Geometric Progression here,

https://brainly.in/question/3619590?msp_poc_exp=5

#SPJ2