A sequence of numbers formed by adding together corresponding terms of an A.Proression and G.Progresion with(common ratio=2).The 1st term =48,2nd term=73,3rd term=128.How to find fourth term? How to write expression (in term n),for nth term of sequence?

2 Answers
WayneDeguMan
Jul 30, 2017

Let the GP be #a, ar, ar^2, ar^3,...#

Let the AP be #b, b + d, b + 2d,...#

so, GP => #a, 2a, 4a, 8a,...#

Adding gives:

a + b = 48...(1)

2a + b + d = 73...(2)

4a + b + 2d = 128...(3)

so, (2) => a + d = 25

and (3) => 3a + 2d = 80

(3) - 2(2) => a = 30

Hence, d = -5 and b = 18

nth term of GP is #ar^(n - 1)#

i.e. #30(2^(n - 1)) = 15(2^n)#

nth term of AP is b + (n - 1)d

i.e. 18 - 5(n - 1) => 23 - 5n

so, nth term of sum is #15(2^n) + 23 - 5n#

Hence, 4th term is #15(2^4) + 23 - 5(4)#

i.e. 15(16) + 23 - 20 => 240 + 23 - 20 = 243

:)>

Tazwar Sikder
Jul 30, 2017

#"Fourth term": u_(4) = 243#

#n"th term": u_(n) = 15 cdot 2^(n) - 5 n + 23#

Explanation:

The sequence of numbers is formed by adding together corresponding terms of an arithmetic progression and a geometric progression.

This means that the first term of the sequence is the sum of the first terms of the #"A.P."# and #"G.P."#, respectively.

The other terms are found in a similar manner.

Let the first term of the #"A.P." = a_(1)# and the first term of the #"G.P." = g_(1)#.

First term: #48 = a_(1) + g_(1)#

Second term: #73 = a_(2) + g_(2)#

Third term: #128 = a_(3) + g_(3)#

Let's express the second and third terms using the #n#th term formulas:

#Rightarrow 73 = (a_(1) + (2 - 1) d) + (g_(1) cdot 2^(2 - 1))#

#Rightarrow 73 = a_(1) + d + 2 g_(1)#

#and#

#Rightarrow 128 = (a_(1) + (3 - 1) d) + (g_(1) cdot 2^(3 - 1))#

#Rightarrow 128 = a_(1) + 2 d + 4 g_(1)#

Then, let's solve both equations for #a_(1)#:

#Rightarrow 73 = a_(1) + d + 2 g_(1)#

#Rightarrow a_(1) = 73 - d - 2 g_(1)#

#and#

#Rightarrow 128 = a_(1) + 2 d + 4 g_(1)#

#Rightarrow a_(1) = 128 - 2 d - 4 g_(1)#

Eliminating #a_(1)# from the two equations:

#Rightarrow 73 - d - 2 g_(1) = 128 - 2 d - 4 g_(1)#

Solving for #d#:

#Rightarrow d = 55 - 2 g_(1)#

Now, let's substitute this expression for #d# back into one of the original equations.

Let's try the first one:

#Rightarrow 73 = a_(1) + (55 - 2 g_(1)) + 2 g_(1)#

#Rightarrow 73 = a_(1) + 55#

#therefore a_(1) = 18#

We now have the value of the first term of the #"A.P."#.

Let's substitute this into the equation for the first term of the sequence:

#Rightarrow 48 = (18) + g_(1)#

#therefore g_(1) = 30#

Before we find the fourth term we must do one more thing.

Earlier we found an expression for the common difference #d#.

Let's substitute the value of #g_(1)# into that expression:

#Rightarrow d = 55 - 2 (30)#

#Rightarrow d = 55 - 60#

#therefore d = - 5#

The fourth term of the sequence will be expressed in the following way:

#Rightarrow u_(4) = a_(4) + g_(4)#

#Rightarrow u_(4) = (a_(1) + (4 - 1) cdot - 5) + (g_(1) cdot 2^(4 - 1))#

#Rightarrow u_(4) = a_(1) - 15 + 8 g_(1)#

#Rightarrow u_(4) = (18) - 15 + 8 (30)#

#Rightarrow u_(4) = 3 + 240#

#therefore u_(4) = 243#

We can find a general expression for the #n#th term of the sequence, as we have now found all relevant values:

#Rightarrow u_(n) = a_(n) + g_(n)#

#Rightarrow u_(n) = (a_(1) + (n - 1) d) + (g_(1) cdot 2^(n - 1))#

#Rightarrow u_(n) = 18 - 5 (n - 1) + frac(30)(2) cdot 2^(n)#

#Rightarrow u_(n) = 18 - 5 n + 5 + 15 cdot 2^(n)#

#therefore u_(n) = 15 cdot 2^(n) - 5 n + 23#

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