Question

the 8th term of the G.P 7,14,28,56,...is ​

Answer 1

EXPLANATION.

8th term of G.P.

Series = 7, 14, 28, 56. . . . .

As we know that,

First term = a = 7.

Common ratio = r = b/a.

Common ratio = r = 14/7 = 2.

Common ratio = r = 2.

As we know that,

Formula of :

General term of G.P.

⇒ Tₙ = a. rⁿ⁻¹.

⇒ T₈ = a. r⁸⁻¹.

⇒ T₈ = a. r⁷.

Put the values in the equation, we get.

⇒ T₈ = 7 x (2)⁷.

⇒ T₈ = 7 x 128.

⇒ T₈ = 896.

                                                                                                                     

MORE INFORMATION.

Supposition of terms in G.P.

(1) = Three terms as : a/r, a, ar.

(2) = Four terms as : a/r³, a/r, ar, ar³.

(3) = Five terms as : a/r², a/r, a, ar, ar²

Answer 2

Answer:

$\small{\orange{ \boxed{ \boxed{ \begin{array}{cc} \hookrightarrow \sf \: \: given \: the \: G.P \: series \: : \\ \\ \sf \: 7,14,28,56. \: . \: . \: \\ \\ \blue{ \underline{\sf \: we \: have \: to \: find : }} \\ \\ \hookrightarrow \sf \: {8}^{th} \: term \: of \: the \: G.P \: series = T_8\\ \\ \\ \red{ \underline{ \sf \: solution :}} \\ \\ \hookrightarrow \: \sf \: first \: term \: of \: the \: G.P \: series, \: a = 7 \\ \\ \hookrightarrow \sf \:common \: ratio \: of \: the \: G.P \: series ,\: r = \frac{14}{2} = 2 \\ \\ \\ \pink{ \underline{ \sf \: we \: know \: that : }} \\ \\ \hookrightarrow \: \sf \: general \: term \: of \:G.P \: series \: \: T_n= a {r}^{n - 1} \\ \\ \\ \bf \: now \: according \: to \: the \: question \: \\ \\ \sf \: T_8 = 7 \times {2}^{8 - 1} \\ \\ = 7 \times {2}^{7} \\ \\ = 896 \\ \\ \blue{ \boxed{\therefore \sf \: {8}^{th} \: term \: of \: the \: series \: is = 896}} \\ \\ \end{array}}}}}$