2.1 Given the following linear series: 3 + 7 + 11 + .. - NSC Mathematics - Question 2 - 2017 - Paper 1

The series can be expressed in sigma notation as:

$\sum_{n=1}^{121} (4n - 1)$

This represents the sum of the terms from $n=1$ to $n=121$, where each term is given by the formula $4n - 1$.

We know that the terms are:

• $T_{10} = 2t - 4$
• $T_{11} = t - 3$
• $T_{12} = 8 - 2t$

For an arithmetic sequence, the difference between consecutive terms is constant. Hence:

$T_{11} - T_{10} = T_{12} - T_{11}$

Substituting the expressions:

$(t - 3) - (2t - 4) = (8 - 2t) - (t - 3)$

Simplifying:

$t - 3 - 2t + 4 = 8 - 2t - t + 3$

Combining like terms:

$-t + 1 = 11 - 3t$

Now, rearranging gives:

We need the value of the first term, which can be determined using the relationship of the terms:

$T_{10} = 2t - 4$

Substituting for $t = 5$ gives:

$T_{10} = 2(5) - 4 = 10 - 4 = 6$

Thus, the first term can be calculated as:

$T_1 = T_{10} - (10-1)d$

Where $d = T_{11} - T_{10} = (t-3) - (2t-4) = -t + 1$. Substituting $t = 5$:

$d = -5 + 1 = -4$

Calculating:

$T_1 = 6 - (9)(-4) = 6 + 36 = 42$

Given the equations:

$T_1 + T_2 = -1$ $T_3 + T_4 = -4$

We start by setting:

$T_n = ar^{n-1}$

From the first equation:

$ar^0 + ar^1 = -1$

This simplifies to: