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
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
20 answers
Sorry, what is the answer to c?
Reiny u're not answering. Please I need the answer now
The third term of a g. P. Is63..and the fifth term is 567. Find the sixth term of the progression. Reiny!!!!!
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
-2,6,-28 . HOW THE COMMON RATIO IS -4
we really appreciate thanks alot,but my comfussion is about question C.
What of (c)
What is the answer for (c)
What is the answer for c in the first question
What is the answer to question (c) I am in desperate need of an answer
Please 🙏🏿 what is the answer for c
What is the answer for c please
I think the answer for c is where Reiny put aum of all terms
6 and 4
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?
and 4Answer for c is 6
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.
The answer for c is 6,4
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.
1 answer