Question

if the 1st term of an infinite geometric series is equal to twice of the sum of all the terms that follows it, then the value of r will be : ?
A)1/3
B)1
C)1/2
D)2
please explain me the answer........!

Answer 1

\[a _{1}=2S _{ \infty }\]
\[S _{ \infty }=a/1-r\]
solve

Answer 2

Can u tell me what U got after solving,.... ?

Answer 3

r=1

Answer 4

Where's the mistake ?

Answer 5

-1<r<1

Answer 6

okay.... BUt the answer on the book is 1/3... I dont know how it comes........... :(

Answer 7

your working is correct,
but the answer cant be -1or 1according to the restriction
the problem is our equations the question says terms that 'follows it'

Answer 8

But @Jonask I here think of something...... U know the sum of infintite geometric series would be applicable only if |r|<1.... but if U got |r|>1. then the formula isn't applicable..... and the question I percept isn't really clear to me... please help me

Answer 9

a and a1 are not the same,
the term a will be
\[a=a _{1}r\]

Answer 10

its \[a _{1},a _{1} r,a _{1}r^2...\]

\[S \infty=a/(1-r)=a _{1}r/1-r\]

Answer 11

the terms following it are
\[a _{1}r,a _{2}r^2,...\]

Answer 12

@Jonask the first term wouldn't be a1r.... Because U know the formula.. \[a_{n}=a_1r ^{n-1}\] so If U will put a=1.... then see what U will obtain.......

Answer 13

these are the terms following (after) a

Answer 14

so I think the series should be like this......
\[a_1,a_1r ^{1},a_1r ^{2}......\]

Answer 15

the series is like that but what about the series for terms after a1

Answer 16

after a...... every number is multiplied by r.... ?

Answer 17

yes
\[a _{1}=2S \infty\]
\[S \infty =a/1-r\]

and \[a=a _{1}r\]

Answer 18

\[a _{1}=2[a/1-r]\]
\[a _{1}=2a _{1}r/1-r\]

Answer 19

\[1-r=2r\]
\[3r=1\]
\[r=1/3\]

Answer 20

So @Jonask in the formula..... what is 'a'?
Its the second term or first......?

Answer 21

@Jonask ........ ?