# Question #7f0e5

Jun 20, 2016

Se explanantion

#### Explanation:

There is no simple answer to this. You must get used to the notation used of which there are slight variances of the letters used depending on preference.

For example
Counting using the letter i to designate position in the sequence and the letter n for the last term position. Another time the letter n could be use instead of i.. So you could have the nth term, ith term or even perhaps using the letter t

Of sequence and series the latter is the hardest (adding up a sequence). For this the only real solution is to find a guide/book with lots of examples and practice questions (with answers).

Sequences follow two basic rules. One for Arithmetic and another for Geometric.

$\textcolor{b l u e}{\text{Arithmetic sequence-basic type:}}$
Let the first value be $a$
Let the difference between terms be $k$
Let the ith term be ${t}_{i}$

${t}_{i} \to {t}_{1} = {a}_{1}$
${t}_{i} \to {t}_{2} = {a}_{1} + k$
${t}_{i} \to {t}_{3} = {a}_{1} + 2 k$
${t}_{i} \to {t}_{4} = {a}_{1} + 3 k$

Notice the pattern!
So for any ${t}_{i} = {a}_{1} + \left(i - 1\right) k$

You can determine $k$ by difference and then determine $a$ by substitution.

You sometimes have alternating positive and negative in which case you would have some variant on:

${t}_{i} = {\left(- 1\right)}^{i} \left[{a}_{1} + \left(i - 1\right) k\right]$
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$\textcolor{b l u e}{\text{Geometric sequence-basic type:}}$

Normally you find people use the letter $r$ for this.

${t}_{i} \to {a}_{1} {r}^{i - 1}$

${t}_{i} \to {t}_{1} = {a}_{1} {r}^{1 - 1} = {a}_{1}$
${t}_{i} \to {t}_{2} = {a}_{1} {r}^{2 - 1} = {a}_{1} r$
${t}_{i} \to {t}_{3} = {a}_{1} {r}^{3 - 1} = {a}_{1} {r}^{2}$
${t}_{i} \to {t}_{4} = {a}_{1} {r}^{4 - 1} = {a}_{1} {r}^{3}$

You can determine r by division.
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$\textcolor{b l u e}{\text{Series}}$

Notation type: ${\sum}_{i = 1 \to n} {a}_{i}$

These you just have to practice as there are many varients.