Question

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

Answer 1

Answer:

$d = q(z-1)$

$S_n = \frac{nq}{2}(2 + (n-1)(z-1))$

$S_{10} = 5q(9z -7)$

Step-by-step explanation:

Given

$T_1 = q$

$T_2 = qz$

Solving (a1): The common difference (d)

d is calculated as

$d = T_2 - T_1$

This gives:

$d = qz - q$

Factorize:

$d = q(z-1)$

Solving (a2): Sum of n terms

This is calculated using:

$S_n = \frac{n}{2}(2*T_1 + (n-1)d)$

Substitute values for T1 and d

$S_n = \frac{n}{2}(2*q + (n-1)q(z-1))$

$S_n = \frac{n}{2}(2q + q(n-1)(z-1))$

Factorize:

$S_n = \frac{n}{2}(q(2 + (n-1)(z-1)))$

$S_n = \frac{nq}{2}(2 + (n-1)(z-1))$

Solving (b): Sum of first 10.

In this case, n = 10

So:

$S_n = \frac{nq}{2}(2 + (n-1)(z-1))$

becomes

$S_{10} = \frac{10 * q}{2}(2 + (10-1)(z-1))$

$S_{10} = 5 * q(2 + 9(z-1))$

$S_{10} = 5q(2 + 9(z-1))$

$S_{10} = 5q(2 + 9z-9)$

$S_{10} = 5q(2 -9+ 9z)$

$S_{10} = 5q(-7+ 9z)$

$S_{10} = 5q(9z -7)$