Let $t_1, t_2, \dots$ be real numbers such that $t_1 + t_2 +\dots+ t_n = 2n^2 + 9n + 13$, for every positive integer $n\ge2$. If $t_k =103$, then $k$ equals

We need to find individual terms from their cumulative sums.

We're told that $t_1 + t_2 + \dots + t_n = 2n^2 + 9n + 13$ for every $n \geq 2$.

Let's call this sum $S_n$, so:

$S_n = t_1 + t_2 + \dots + t_n = 2n^2 + 9n + 13$

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Here's the key insight: If we know the sum of the first $n$ terms and the sum of the first $(n-1)$ terms, we can find the $n$-th term by subtraction.

$t_n = S_n - S_{n-1}$

Let's calculate a few terms:

For $n = 2$: $S_2 = 2(2^2) + 9(2) + 13 = 8 + 18 + 13 = 39$

For $n = 3$: $S_3 = 2(3^2) + 9(3) + 13 = 18 + 27 + 13 = 58$

Therefore: $t_3 = S_3 - S_2 = 58 - 39 = 19$

For $n = 4$: $S_4 = 2(4^2) + 9(4) + 13 = 32 + 36 + 13 = 81$

Therefore: $t_4 = S_4 - S_3 = 81 - 58 = 23$

For $n = 5$: $S_5 = 2(5^2) + 9(5) + 13 = 50 + 45 + 13 = 108$

Therefore: $t_5 = S_5 - S_4 = 108 - 81 = 27$

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We look at the terms we found: $t_3 = 19$, $t_4 = 23$, $t_5 = 27$

The differences are: $23 - 19 = 4$ and $27 - 23 = 4$

This means the terms form an arithmetic progression (AP) starting from $t_3$!

An arithmetic progression is a sequence where each term differs from the previous by a constant value (called the common difference).

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For an AP with first term $a$ and common difference $d$, the $n$-th term is: $a + (n-1)d$

In our case, starting from $t_3$:

First term: $t_3 = 19$

Common difference: $d = 4$

For the $n$-th term where $n \geq 3$: $t_n = 19 + (n-3) \times 4$

Simplifying: $t_n = 19 + 4n - 12 = 4n + 7$

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We need to find $k$ such that $t_k = 103$.

Since our formula works for $n \geq 3$, and $103 > 19 = t_3$, we know $k \geq 3$.

Using our formula: $t_k = 4k + 7 = 103$

$4k = 103 - 7 = 96$

Therefore: $k = 24$

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Let's check: $t_{24} = 4(24) + 7 = 96 + 7 = 103$

Answer: $k = 24$