Question

For the G.P. a = 6,r = 2
then S10=​

Answer 1

EXPLANATION.

In the G.P.

⇒ a = 6.

⇒ r = 2.

As we know that,

First term of a G.P. = a = 6.

Common ratio of a G.P. = r = 2.

As we know that,

Formula of :

⇒ Sₙ = a(rⁿ - 1)/r - 1.

⇒ S₁₀ = 6(2¹⁰ - 1)/2 - 1.

⇒ S₁₀ = 6(2¹⁰ - 1).

⇒ S₁₀ = 6(1024 - 1).

⇒ S₁₀ = 6(1023).

⇒ S₁₀ = 6138.

                                                                                                                           

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

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${\pmb{\sf{\underline{Explaination...}}}}$

$\begin{gathered}\boxed{\begin{array}{c}\\ \bf \bigstar \: Required \; Solution \\ \\ \star \: \sf{There \: is \: GP \: (Geometric \: progression) \: given} \\ \\ \star\sf \: {We \: have \: to \: find \: S_{10}} \\ \\ \star \sf \: Here, \begin{cases} & \sf{r \: is \: \bf{2}} \\ & \sf{a \: is \: \bf{6}} \end{cases} \end{array}}\end{gathered}$

${\pmb{\sf{\underline{Using \; formula...}}}}$

${\small{\underline{\boxed{\sf{S_n \: = a(r^n - 1)/r-1}}}}}$

${\pmb{\sf{\underline{Full \; Solution...}}}}$

$\sf \underset{\sf Using \; Formula}{\underbrace{\small{\boxed{\pink{\sf{S_n \: = a(r^n - 1)/r-1}}}}}} \\ \\ :\implies \sf S_n \: = a(r^n - 1)/r-1 \\ \\ :\implies \sf S_{10} \: = 6(2^{10} - 1)/2-1 \\ \\ :\implies \sf S_{10} \: = 6(2^{10} - 1)/1 \\ \\ :\implies \sf S_{10} \: = 6(2^{10} - 1) \\ \\ :\implies \sf S_{10} \: = 6(1024 -1) \\ \\ :\implies \sf S_{10} \: = 6(1023) \\ \\ :\implies \sf S_{10} \: = 6 \times 1023 \\ \\ :\implies \sf S_{10} \: = 6138$

Henceforth, ${\bf{S_{10}}}$ is 6138.