Which term of the GP 5,10,20,40,...........is 5120?

G.S.Kantharao   8 years AGO

G.P=5,10,20.....5120
n th term = a* r^(n-1) = 5120 Here a= 5, r= 2
5*2^(n-1) = 5120
2^(n-1) = 5120/5 = 1024
2^(n-1) = 2^10
n = 11. Hence the right option is 6.

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Ajit Chaudhary   8 years AGO

Here, a=5, r=2, t(n) = 5120
Since, we know
t(n) = a*r^(n-1)
5120 = 5*(2)^(n-1)
by solving we will get n=11
So 11th element will be 5120 in the given GP
Option (6) correct

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Shavy khanna   8 years AGO

Each next term is double of its previous term.following this 5120 is the 11th term.Option 6 is right answer.

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Shefali jain   8 years AGO

a1 = 5
r= 2
an = a1 * (r)^n-1
5120 = 5 * (2)^n-1
(2)^10 = (2)^n-1
-> n=11
Hence 11th term will be 5120
Hence Option 6 is correct

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lakshmi nalluri   8 years AGO

a.r^n-1/r-1 =5120
a=5,r=2 on substituting,
n=11

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Jasmin Ara   8 years AGO

10/5=2, which is 2^1, term is 2nd term,
20/5=4 which is 2^2, term is 3rd term, (power of the 2 +1)=3
40/5=8,which is 2^3, term is 4th term, (power of the 2 +1)=4
5120/5=1024, 2^10, term is 11th term.

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gubba ranjith kumar   8 years AGO

5.r^(n-1)=5120
5.2^(n-1)=5120
=>2^(n-1)=1024
=>n-1=12 =>n=13

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Samjukta Das   8 years AGO

5*2=10
10*2=20
20*2=40
40*2=80
80*2=160
160*2=320
320*2=640
640*2=1280
1280*2=2560
2560*2=5120 i.e. 11th term will be 5120

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Dinesh Sharma   8 years AGO

Last Term = First Term * (ratio)^(no. of terms - 1)
i.e. 5120 = 5*(2)^(n-1) where r = 2nd Term/1st
i.e. term=10/5=2
i.e. 5120/5 = 2^(n-1)
i.e. 1024 = 2^(n-1)
i.e. 2^10 = 2^(n-1)
i.e. n-1=10
i.e. n=11th term

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durga prasad   8 years AGO

=>nth term = a1*r^(n-1)
=>5120=5*2^(n-1)
=>1024=2^(n-1)
=>2^10=2^(n-1)
=>10=n-1
=>n=11th

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Venkatesh kadali   8 years AGO

common ratio r=second term/first term
r=10/5
r=2



nth term = first term * (r)^n-1
5120=5 * (2)^n-1
1024=2^n-1
2^10=2^n-1
10=n-1
n=11
so answer is 6 option

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harsh dhir   8 years AGO

ar^n-1=5120
a=5
r=2
Therefore n=11

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Pragati Rode   8 years AGO

To find the nth term use the formula is ar^n-1

so using the formula

where a= 5 and r = 2 and nth term is 5120

5120 = 5 * 2^n-1
1024 = 2^n-1
2^10 = 2^n-1
10 = n-1
n= 11
so the 5120 is the 11th term of gp

so option is 6

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shubham   8 years AGO

a (r^(n-1))=5120
a=5
r=2
n will equal to 11

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Nitish kesarwani   8 years AGO

tn= ar^(n-1) where tn=5120
a=5 c.r=2
using formula 5120/5=(2)^(n-1)
taking log
2^10=(2)^(n-1)
10=n-1 so n=11th term option 6

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Parnab Ghosh   8 years AGO

a*r^(n-1)=5120 [a=5,r=2]
=>5*2^(n-1)=5120
=>2^(n-1)=5120/5
=>2^(n-1)=1024
=>2^(n-1)=2^10
so, n-1=10 [Since base is equal]
n=11(Ans)

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Vijay Kumar   8 years AGO

1st term=5
2nd term=5*2=10
3rd =5*4=20
4th = 5*8=40
5th = 5*16=80
6th =5*32=160
7th = 5*64=320
8th = 5*128=640
9th = 5*256=1280
10th = 5*512=2560
11th = 5*1024=5120.
so,The given no 5120 is 11th term of this series.

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Siva   8 years AGO

a×(r^(n-1))=5120
2^(n-1) =5210÷5=1024
n-1 =10
n=11

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sowmya   8 years AGO

5120= 5*(2^(n-1))

10240/5= 2^n
2048=2^n
2^11=2^n, hence
n=11

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Parikshit Sen   8 years AGO

In a g.p series,
tn=ar^n-1.
atq,
5120=5*2^n-1
or, n=11

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Vijaya mounika nallapusa   8 years AGO

a=5,ar=10
So r=2
ar^(n-1)=5120
r^(n-1)=1024
n=11

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anubhav verma   8 years AGO

5120 = a* r^(n-1)
a=st term=5 , n= nth term , r= factor=2

5120 = 5* 2^(n-1)
1024 = 2^(n-1)
2^10 = 2^(n-1)
n-1 =10
n=1

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codemaker   8 years AGO

5*(2)^(n-1)=5120
2^(n-1)=5120/5
2^(n-1)=1024
n-1=10
n=11

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YASWANTH REDDY   8 years AGO

here, a=5, r=2. then let 5120 be the n'th term.
therefore, a*r^(n-1) = 5120.
solving this, we get n=11

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ANAND MOHAN RAI   8 years AGO

a.(r)^(n-1)=5120
here a=5 and r=2
hence 5.(2)^n-1=5120
n=11
hence option 6:)

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P.V.Vishnu Prasanth   8 years AGO

nth term=ar^n-1
a=5, r=2
5120=5(2^n-1); 1024=2^n-1; 2^10=2^n-1; n=11

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Siddartha Mulumudi   8 years AGO

11th term.

Common Ratio is 2, 1st term is 5

nth term is ar^(n-1) =5120. Solve the above and we get n as 11

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Gurpreet Kaur   8 years AGO

5,10,20,40,80,160,320,640,1280,2560,5120...
5120 is11th term in the series

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Avinash Singh   8 years AGO

5120=5*(2)^(n-1)
1024=2^(n-1)=2^10
n-1=10
n=11
option 6) 11th

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Venkatesh Rajendran   8 years AGO

nth term = ar^(n-1)
5120=5*2^(n-1)
1024=2^(n-1)
2^10=2^(n-1)
10=n-1
n=11

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Prathapreddy   8 years AGO

5, 5×2,10×2,20×2,40×2,80×2,160×2........2510×2=5120 (11th)

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SATHEESH V   8 years AGO

a=5
r=10/5=2
a(r^(n-1))=5120
r^(n-1)=1024

therefore n=11

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Nandini JB   8 years AGO

an=a1*r^(n-1)
5120=5*2^(n-1)
1024=2^(n-1)
2^10=2^(n-1)
n=11

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shamil muhammed   8 years AGO

for a Gp nth element is 'ar^(n-1)
where 'a' is first element ,'r' be the common ratio.
Here a=5,r=2 then
5*2^(n-1)=5120
2^(n-1)=1024
then
n=11
so option 6

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THONDURU PRADEEP KUMAR   8 years AGO

a1=5,r=2;
an=a1q^n-1=>5120=5*2^n-1
=>5120=5*2^n*2^-1
=>5120=5/2*2^n
=>5120*2/5=2^n
=>2048=2^n
=>2^11=2^n
=>n=11

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Thatikonda Saikrishna   8 years AGO

By multiplying the previous number by 2 we get 5120 then counting from 1st it would be 11th term.

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divyasree   8 years AGO

here first term a=2
diff r = 2. (10/5)
5120 is ar^(n-1) term
5×2^(n-1)=5120
2^(n-1)=5120÷5
2^(n-1)=1024
2^(n-1)=2^10
(n-1)=10
n=11

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brajabijayini acharya   8 years AGO

its a g.p for sure
5*2=10 10*2=20
so here a=5, r=2,
5*2^n-1=5120
2^n-1=1024
2^n-1=2^10
n-1=10
n=11

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