Question

for the given progression find the 9th term 2,6,8,54...​

Answer 1

Correct Question :--- Find the 9th term of progression :-- 2,6,18,54 ____________

Concept used :---

$\bigstar\underline{\red{\mathbb{AP\:\green{SERIES}}}}\bigstar$

If in an A.P. series "a" be the first term and "d" be the common difference then ,

(1) The n'th term is given by the formula .

$\longrightarrow \red{\boxed{\sf a_n=a+(n-1)d}}$

(2)Sum of n number of terms ,

$\longrightarrow\green{\boxed{\sf S_n=\frac{n[2a+(n-1)d]}{2}}}$

━━━━━━━━━━━━━━━━━━━━━━━

$\bigstar\underline{\red{\mathbb{GP\:\pink{SERIES}}}}\bigstar$

If in an G.P. series "a" be the first term and "r" be the common ratio then ,

(1) The n'th term is given by the formula .

$\longrightarrow\pink{\boxed{\tt \: a_n=at{}^{n-1}}}$

(2)Sum of n number of terms ,

$\longrightarrow \purple{\boxed{\rm \: S_n=\frac{n[r{}^{n}-1]}{r-1}}}$

━━━━━━━━━━━━━━━━━━━━━━━

$\underline {\underline{\LARGE{{\bf{\color{lime}{S}}}{\mathfrak{o}}{\mathfrak{\orange{l}}}{\mathfrak{\color{aqua}{u}}}{\mathfrak{\pink{t}}}{\mathfrak{\purple{i}}}{\mathfrak{\color{navy}{o}}}{\mathfrak{\red{n}}}}}} :$

$\green{\tt \: Given \: Series \: is = 2 \: 6 \: 18 \: 54 - - -} \\ \\ \red\longmapsto\:\rm \: first \: term(a) = 2 \\ \\ \red\longmapsto\:\rm \: common \: ratio(r) = \frac{6}{2} = 3 \\ \\ \blue{\bf \: putting \: values \: now} \: \\ \\ \bf \: T_9 \: = a \times r^{9 - 1} \\ \\ \red\longmapsto\:\rm \: T_9 = 2 \times {3}^{8} \\ \\ \red\longmapsto\: \pink{\boxed{\large\rm \red{T_9} \green= \purple{13122}}}$

___________________________

$\red{\textbf{Hence, values of 9th Term of GP is 13122..}}$