Question

Which term from end of the progression 3, 6, 12, ... , 3072 is 384.​

Answer 1

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

GIVEN :

Progression = 3, 6, 12, ... , 3072.

TO FIND :

The term which is 384 from last.

SOLUTION :

If the analysis the given progression we can identify that it is in G.P where

l = 3072, r = 2, a(n) = 384.

Now,

$\implies \: \sf{a_{n} = \frac{l}{ {r}^{n - 1} }}$ [Formula used to find last term from the end]

On putting the values we get,

$\sf {\implies \: 384 = \frac{3027}{ {(2)}^{n - 1}}}$

$\sf {\implies \: {(2)}^{n - 1} = \frac{3072}{384} }$

$\sf {\implies \: {(2)}^{n - 1} = 8}$

$\sf {\implies \: {(2)}^{n - 1} = {2}^{3} }$

$\implies \: \sf{n - 1 = 3}$

$\sf{ \implies \: n = 4}$

$\boxed {\sf{Hence, \: the \: required \: term \: is \: 4}}$

Answer 2

Answer:

$GIVEN :</p><p>Progression = 3, 6, 12, ... , 3072.</p><p></p><p>TO FIND :</p><p>The term which is 384 from last.</p><p></p><p>SOLUTION :</p><p>If the analysis the given progression we can identify that it is in G.P where</p><p></p><p>l = 3072, r = 2, a(n) = 384.</p><p></p><p>Now,</p><p></p><p>\implies \: \sf{a_{n} = \frac{l}{ {r}^{n - 1} }}⟹a </p><p>n</p><p> </p><p> = </p><p>r </p><p>n−1</p><p> </p><p>l</p><p> </p><p> [Formula used to find last term from the end]</p><p></p><p>On putting the values we get,</p><p></p><p>\sf {\implies \: 384 = \frac{3027}{ {(2)}^{n - 1}}}⟹384= </p><p>(2) </p><p>n−1</p><p> </p><p>3027</p><p> </p><p> </p><p></p><p>\sf {\implies \: {(2)}^{n - 1} = \frac{3072}{384} }⟹(2) </p><p>n−1</p><p> = </p><p>384</p><p>3072</p><p> </p><p> </p><p></p><p>\sf {\implies \: {(2)}^{n - 1} = 8}⟹(2) </p><p>n−1</p><p> =8</p><p></p><p>\sf {\implies \: {(2)}^{n - 1} = {2}^{3} }⟹(2) </p><p>n−1</p><p> =2 </p><p>3</p><p> </p><p></p><p>\implies \: \sf{n - 1 = 3}⟹n−1=3</p><p></p><p>\sf{ \implies \: n = 4}⟹n=4</p><p></p><p>\boxed {\sf{Hence, \: the \: required \: term \: is \: 4}} </p><p>Hence,therequiredtermis4</p><p>$