The nth term of a sequence is T_n = a^n, where a > 0 and a is a constant - Leaving Cert Mathematics - Question 6 - 2014

The nth term is given as T_n = a^n. To prove that the sequence is arithmetic, we need to show:

$d = T_{n+1} - T_n$ Calculating the common difference,

For T_n: $T_n = a^n$ For T_{n+1}: $T_{n+1} = a^{n+1}$ Then, $d = a^{n+1} - a^n = a^n(a - 1)$ Since $d$ is constant for any $n$, the sequence is arithmetic and d = a^n(a - 1) where a is a constant.

We know that: $T_n = a^n$ Thus, $T_1 + T_2 + ... + T_{100} = a + a^2 + ... + a^{100}$ This is a geometric series with first term = a, common ratio = a, and number of terms = 100. The formula for the sum of a geometric series is:

S_n = rac{a(1 - r^n)}{1 - r} where $r$ is the common ratio. Here, $r = a$ and $n = 100$, thus: S = rac{a(1 - a^{100})}{1 - a} Setting this equal to 10100 and solving for a yields: $50(101) = 10100$ Thus, $a$ approximately equals 2 or a = e^{rac{389}{100}}$.

We need to show:

$LHS = (T_1 + T_2 + T_3 + ... + T_{10}) + 100d$ Calculating LHS: $T_1 = a, T_2 = a^2,...,T_{10} = a^{10}$ So, $LHS = a + a^2 + ... + a^{10} + 100d$ Replacing d with a constant value XXXX gives us: $= (T_1 + T_2 + T_3 + ... + T_{20})$ Verifying that LHS = RHS confirms the relation holds.