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Given that for an arithmetic sequence, the first term is q and the second term is qz.

(a) Find the common difference d the general term Tn and the sum of the first n terms. Answer must

be in terms of q and z.
(b) Hence, find the sum of the first 10 terms in terms of q and z. ​

Respuesta :

MrRoyal

Answer:

[tex]d = q(z-1)[/tex]

[tex]S_n = \frac{nq}{2}(2 + (n-1)(z-1))[/tex]

[tex]S_{10} = 5q(9z -7)[/tex]

Step-by-step explanation:

Given

[tex]T_1 = q[/tex]

[tex]T_2 = qz[/tex]

Solving (a1): The common difference (d)

d is calculated as

[tex]d = T_2 - T_1[/tex]

This gives:

[tex]d = qz - q[/tex]

Factorize:

[tex]d = q(z-1)[/tex]

Solving (a2): Sum of n terms

This is calculated using:

[tex]S_n = \frac{n}{2}(2*T_1 + (n-1)d)[/tex]

Substitute values for T1 and d

[tex]S_n = \frac{n}{2}(2*q + (n-1)q(z-1))[/tex]

[tex]S_n = \frac{n}{2}(2q + q(n-1)(z-1))[/tex]

Factorize:

[tex]S_n = \frac{n}{2}(q(2 + (n-1)(z-1)))[/tex]

[tex]S_n = \frac{nq}{2}(2 + (n-1)(z-1))[/tex]

Solving (b): Sum of first 10.

In this case, n = 10

So:

[tex]S_n = \frac{nq}{2}(2 + (n-1)(z-1))[/tex]

becomes

[tex]S_{10} = \frac{10 * q}{2}(2 + (10-1)(z-1))[/tex]

[tex]S_{10} = 5 * q(2 + 9(z-1))[/tex]

[tex]S_{10} = 5q(2 + 9(z-1))[/tex]

[tex]S_{10} = 5q(2 + 9z-9)[/tex]

[tex]S_{10} = 5q(2 -9+ 9z)[/tex]

[tex]S_{10} = 5q(-7+ 9z)[/tex]

[tex]S_{10} = 5q(9z -7)[/tex]

akposevictor

Given that the first term and the second term of an arithmetic sequence is q and qz respectively, therefore:

a. Common difference (d) = [tex]\mathbf{d = q(z - 1)}[/tex]

Sum of the first n terms is: [tex]\mathbf{S_n = \frac{qn}{2}[2 + (n - 1)(z - 1)]}[/tex]

b. [tex]\mathbf{S_{10} = 5q(9z - 7)}[/tex]

Recall:

  • The common difference (d) of an arithmetic sequence = difference between the next term and the previous term
  • Sum of n terms, [tex]S_n[/tex] of arithmetic sequence = [tex]\frac{n}{2}[2a + (n - 1)d][/tex]

Given:

  • First term, [tex]T_1 = q[/tex] (this is also "a")
  • Second term, [tex]\\\\T_2 = qz[/tex]

a. Common difference (d) = [tex]T_2 - T_1[/tex]

  • Substitute

[tex]d = qz - q\\\\\mathbf{d = q(z - 1)}[/tex]

Sum of n terms:

[tex]T_n = \frac{n}{2}[2a + (n - 1)d][/tex]

  • Substitute by plugging in the value of d and a

[tex]S_n = \frac{n}{2}[2(q) + (n - 1)q(z - 1)]\\\\S_n = \frac{n}{2}[2q + q(n - 1)(z - 1)]\\\\S_n = \frac{qn}{2}[2 + 1(n - 1)(z - 1)]\\\\\mathbf{S_n = \frac{qn}{2}[2 + (n - 1)(z - 1)]}[/tex]

b. Sum of the first 10 terms:

  • Substitute n = 10 into  [tex]S_n = \frac{qn}{2}[2 + (n - 1)(z - 1)][/tex]

[tex]S_{10} = \frac{q(10)}{2}[2 + (10 - 1)(z - 1)][/tex]

  • Simplify

[tex]S_{10} = 5q[2 + (9)(z - 1)]\\\\S_{10} = 5q(2 + 9z - 9)\\\\[/tex]

  • Add like terms

[tex]\mathbf{S_{10} = 5q(9z - 7)}[/tex]

In summary, given that the first term and the second term of an arithmetic sequence is q and qz respectively, therefore:

a. Common difference (d) = [tex]\mathbf{d = q(z - 1)}[/tex]

Sum of the first n terms is: [tex]\mathbf{S_n = \frac{qn}{2}[2 + (n - 1)(z - 1)]}[/tex]

b. [tex]\mathbf{S_{10} = 5q(9z - 7)}[/tex]

Learn more here:

https://brainly.com/question/14311840

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