Asked by Majors
The first and second terms of an exponential sequence (G.P) are respectively the first and third terms of a linear sequence (A.P). The fourth term of the linear sequence is 10 and sum of its first five terms is 60. find (a) the first five terms of the linear sequence and the sum of the first n terms. (b) the sum Sn of the first n terms of the exponential sequence. (c) the limit of Sn for large value of n
Answers
Answered by
Reiny
let the GP be
a , ar, ar^2 , ar^3, ...
the arithmetic (linear) sequence is
a , a+d , a+2d , a + 3d , a+4d ..
but 2nd term of GP= 3rd term of As
ar = a + 2d *
also for the AP
a+3d = 10 **
and 5a + 10d = 60
a + 2d = 12 ***
subtract (**) - (***)
d = -2 <------------
then in **
a - 6 = 10
a = 16 <------------
now back in *
16r = 16 -4
r = 12/16 = 3/4 <-----------
the GP is
16 , 12 , 9 , 27/4 , ...
a)
the AP is
16, 14 , 12 , 10 , 8 ,6 ...
sum(n) = (n/2)(2a + (n-1)d)
= (n/2)(32 - 2(n-1))
= (n/2)(34 - 2n)
= n(17 - n)
b) for the GP
sum(n) = a(1-r^n)/(1-r)
= 16( 1 - (3/4)^n)/(1-3/4)
= 64(1 - (3/4)^n)
sum(all terms) = a/(1-r)
= 16/(1-3/4)
= 64
a , ar, ar^2 , ar^3, ...
the arithmetic (linear) sequence is
a , a+d , a+2d , a + 3d , a+4d ..
but 2nd term of GP= 3rd term of As
ar = a + 2d *
also for the AP
a+3d = 10 **
and 5a + 10d = 60
a + 2d = 12 ***
subtract (**) - (***)
d = -2 <------------
then in **
a - 6 = 10
a = 16 <------------
now back in *
16r = 16 -4
r = 12/16 = 3/4 <-----------
the GP is
16 , 12 , 9 , 27/4 , ...
a)
the AP is
16, 14 , 12 , 10 , 8 ,6 ...
sum(n) = (n/2)(2a + (n-1)d)
= (n/2)(32 - 2(n-1))
= (n/2)(34 - 2n)
= n(17 - n)
b) for the GP
sum(n) = a(1-r^n)/(1-r)
= 16( 1 - (3/4)^n)/(1-3/4)
= 64(1 - (3/4)^n)
sum(all terms) = a/(1-r)
= 16/(1-3/4)
= 64
Answered by
T. Jay
Sorry, what is the answer to c?
Answered by
T. Jay
Reiny u're not answering. Please I need the answer now
Answered by
T. Jay
The third term of a g. P. Is63..and the fifth term is 567. Find the sixth term of the progression. Reiny!!!!!
Answered by
jane
T3=ar^2=63 ...i
T5=ar^4=567 ...ii
diving ii by i
ar^4/ar^2=567/63
r^2=9
r=3
ar^2=63
a=63/9
a=7
T6=ar^5
=7*243
=1701
T5=ar^4=567 ...ii
diving ii by i
ar^4/ar^2=567/63
r^2=9
r=3
ar^2=63
a=63/9
a=7
T6=ar^5
=7*243
=1701
Answered by
SULAIMAN ABUBAKAR
-2,6,-28 . HOW THE COMMON RATIO IS -4
Answered by
uzairu abubakar
we really appreciate thanks alot,but my comfussion is about question C.
Answered by
Elizabeth
What of (c)
Answered by
Abimbola Salami
What is the answer for (c)
Answered by
Ike
What is the answer for c in the first question
Answered by
Dennis
What is the answer to question (c) I am in desperate need of an answer
Answered by
Olamide
Please 🙏🏿 what is the answer for c
Answered by
Joel
What is the answer for c please
Answer
I think the answer for c is where Reiny put aum of all terms
Answered by
Star boy
6 and 4
Answered by
Star boy
and 4Answer for c is 6
Answered by
Star boy
The answer for c is 6,4
Answered by
Bot
I'm sorry, I cannot determine which question you are referring to. Can you please provide more context and information so I can assist you better?
Answered by
Bot
I apologize for any confusion earlier. Yes, the sum of the first n terms for the exponential sequence is given by:
sum(n) = a(1-r^n)/(1-r)
For large values of n, as r is less than 1, r^n becomes very small, tending towards 0. Therefore, the sum of the first n terms approaches a/(1-r).
In this case, the sum of all terms in the exponential sequence is:
sum(all terms) = a/(1-r) = 16/(1-3/4) = 64
Hence, the answer to part (c) is 64.
sum(n) = a(1-r^n)/(1-r)
For large values of n, as r is less than 1, r^n becomes very small, tending towards 0. Therefore, the sum of the first n terms approaches a/(1-r).
In this case, the sum of all terms in the exponential sequence is:
sum(all terms) = a/(1-r) = 16/(1-3/4) = 64
Hence, the answer to part (c) is 64.
Answered by
Bot
I apologize for the error in my earlier response.
To find the limit of the sum of the first n terms as n approaches infinity, we need to consider the value of the common ratio (r) for the exponential sequence.
In this case, r = 3/4, which is less than 1. Therefore, as n approaches infinity, r^n becomes smaller and smaller, approaching 0. Hence, the sum of the first n terms approaches a finite value equal to:
a/(1 - r)
Substituting the given values, we get:
a/(1 - r) = 16/(1 - 3/4) = 64
Therefore, the limit of the sum of the first n terms as n approaches infinity is 64.
Again, I apologize for any confusion my earlier response may have caused.
To find the limit of the sum of the first n terms as n approaches infinity, we need to consider the value of the common ratio (r) for the exponential sequence.
In this case, r = 3/4, which is less than 1. Therefore, as n approaches infinity, r^n becomes smaller and smaller, approaching 0. Hence, the sum of the first n terms approaches a finite value equal to:
a/(1 - r)
Substituting the given values, we get:
a/(1 - r) = 16/(1 - 3/4) = 64
Therefore, the limit of the sum of the first n terms as n approaches infinity is 64.
Again, I apologize for any confusion my earlier response may have caused.