find the 6th term of an arithmetic sequence with a1=1,a11=54

-103

Step-by-step explanation:

a(1) = 12

a(n) = a(n-1) - 23

a(2) = a(1) - 23

       = 12 - 23

      = -11

Common difference (d) = a(2) -a(1) = -11 - 12  = -23

a(n) =  a1 + (n-1)d

a(6) = 12 + 5 * (-23)

       = 12 - 115

       = -103

-103

Step-by-step explanation:

a(1) = 12

a(n) = a(n-1) - 23

a(2) = a(1) - 23

       = 12 - 23

      = -11

Common difference (d) = a(2) -a(1) = -11 - 12  = -23

a(n) =  a1 + (n-1)d

a(6) = 12 + 5 * (-23)

       = 12 - 115

       = -103

56

Step-by-step explanation:

The n th term of an AP is

= a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₆ = 17 and a₁₃ = 38, then

a₁ + 5d = 17 → (1)

a₁ + 12d = 38 → (2)

Subtract (1) from (2) term by term to eliminate a₁

7d = 21 ( divide both sides by 7 )

d = 3

Substitute d = 3 into (1) and evaluate for a₁

a₁ + 5(3) = 17

a₁ + 15 = 17 ( subtract 15 from both sides )

a₁ = 2

Thus

= 2 + (18 × 3) = 2 + 54 = 56

20th term= 119

Step-by-step explanation:

Let the first term of AP be a

and common difference be d

nth term is given by,

= a+(n-1)d

6th term = 35

a+(6-1)d= 35

a+5d=35 (i)

13th term = 77

a+12d= 77(ii)

Now, equation (ii) - (i)

a+12d-(a+5d)= 77-35

a+12d-a-5d= 42

7d=42

d=6

Now, from equation (i)

a= 35-5d= 35-5(6)= 35-30 = 5

20th term= a+19d= 5+19(6)= 119

56

Step-by-step explanation:

The n th term of an AP is

= a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₆ = 17 and a₁₃ = 38, then

a₁ + 5d = 17 → (1)

a₁ + 12d = 38 → (2)

Subtract (1) from (2) term by term to eliminate a₁

7d = 21 ( divide both sides by 7 )

d = 3

Substitute d = 3 into (1) and evaluate for a₁

a₁ + 5(3) = 17

a₁ + 15 = 17 ( subtract 15 from both sides )

a₁ = 2

Thus

= 2 + (18 × 3) = 2 + 54 = 56

5

Step-by-step explanation:

The 6th term of an arithmetic progression is 35 and the 13th term is 77. find the 1st term

We know we have an arithmetic sequence.

a_n = (n - 1)*d + a_1

given:

a_6 = 35 = (6-1)*d + a_1

a_13 = (13 -1)*d + a_1 = 77

find a_1

we have  35 = 5d + a_1

and  77 = 12d + a_1

2 equations in 2 unknowns.

We can solve this.

77 = 12d + a_1

35 = 5d + a_1

minus  the 2 equations from each other.

77 - 35 = 12d - 5d + a_1 - a_1

42 = 7d

d = 6

Find a_1

35 = 5*6 + a_1

35 = 30 + a_1

a_1 = 35 -30 = 5

a_1 = 5

20th term= 119

Step-by-step explanation:

Let the first term of AP be a

and common difference be d

nth term is given by,

= a+(n-1)d

6th term = 35

a+(6-1)d= 35

a+5d=35 (i)

13th term = 77

a+12d= 77(ii)

Now, equation (ii) - (i)

a+12d-(a+5d)= 77-35

a+12d-a-5d= 42

7d=42

d=6

Now, from equation (i)

a= 35-5d= 35-5(6)= 35-30 = 5

20th term= a+19d= 5+19(6)= 119

16

Step-by-step explanation:

Let's set the arithmetic : -4,0,4,8,x,y,...

The number we need to find is y

We have:

-4 + 4 = 0

0 + 4 = 4

4 + 4 = 8

So:

x = 8 + 4 = 12

y = x + 4 = 12 + 4 = 16

Then the 6th term of the arithmetic is 16 !

Brainliest, please?

a(first term)=4

d(common difference)=6

sum of first 6 terms=114