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BabaBlast [244]
3 years ago
13

two terms of an arithmetic sequence are a6= 40 and a20= -16. write and explicit rule for the nth term

Mathematics
1 answer:
Lana71 [14]3 years ago
7 0

Answer:

Tn = 64-4n

Step-by-step explanation:

The nth term of an AP is expressed as:

Tn = a+(n-1)d

a is the common difference

n is the number of terms

d is the common difference

Given the 6th term a6 = 40

T6 = a+(6-1)d

T6 = a+5d

40 = a+5d ... (1)

Given the 20th term a20 = -16

T20 = a+(20-1)d

T20 = a+19d

-16 = a+19d... (2)

Solving both equation simultaneously

40 = a+5d

-16 = a+19d

Subtracting both equation

40-(-16) = 5d-19d

56 = -14d

d = 56/-14

d = -4

Substituting d = -4 into equation

a+5d = 40

a+5(-4) = 40

a-20 = 40

a = 20+40

a = 60

Given a = 60, d = -4, to get the nth term of the sequence:

Tn = a+(n-1)d

Tn = 60+(n-1)(-4)

Tn = 60+(-4n+4)

Tn = 60-4n+4

Tn = 64-4n

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yan [13]

Answer:

 x = -2/5

 y = 12/5

Step-by-step explanation:

Substitution method;

-2x + 3y = 8  -----------(i)

x + y =2   -----------(ii)

x = 2 - y

substitute the value of x in equ (i)

  (-2)* (2-y) + 3y = 8

     -4 + 2y + 3y = 8

                   5y = 8 + 4

                     y = 12/5

substitute the value of y in equ (ii)

x + 12/5 = 2

        x = 2 - 12/5

           = 2*5/1*5 - 12/5  

          = 10 -12/5 = -2/5

Ans: x = -2/5

       y = 12/5

elimination method:

-2x + 3y = 8  -----------(i)

x + y =2   -----------(ii)

                                    (i)   ====>    -2x + 3y = 8

multiply equ (ii) by 2         ====>    <u>  2x + 2y = 4</u>

             add (i) and (ii)                             5y = 12

                                                                   y = 12/5

Substitute in equ (ii)

x + 12/5 = 2

        x = 2 - 12/5

           = 2*5/1*5 - 12/5  

          = 10 -12/5 = -2/5

Ans: x = -2/5

       y = 12/5

                                                       

               

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Step-by-step explanation:

Given expression:

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Problem;

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