Answer:
5th term is -25
and sum of first three terms is -21
Step-by-step explanation:
We are given the sequence:
[tex] \displaystyle \large{a_n = 5 - 6n}[/tex]
To find 5th term, substitute n = 5.
[tex] \displaystyle \large{a_5 = 5 - 6(5)} \\ \displaystyle \large{a_5 = 5 - 30} \\ \displaystyle \large{a_5 = - 25}[/tex]
Therefore, fifth term is 25.
Next, to find the sum of first three terms, we will introduce sigma.
[tex] \displaystyle \large{a_1 + a_2 + a_3 + ... + a_n = \sum_{k = 1}^{n} a_k}[/tex]
Our ak is 5-6k
Since we want to find sum of first three terms:-
[tex] \displaystyle \large{ \sum_{k = 1}^{3}( 5 - 6k)}[/tex]
Expand Sigma in.
[tex] \displaystyle \large{ \sum_{k = 1}^{3} 5 + \sum_{k = 1}^{3}- 6k}[/tex]
Property of Summation
[tex] \displaystyle \large{ \sum_{k = 1}^{n} m = m \times n \: \: \: \sf{(m \: \: is \: \: constant})} \\ \displaystyle \large{\sum_{k = 1}^{n}(a_k + b_k) = \sum_{k = 1}^{n}a_k + \sum_{k = 1}^{n}b_k} \\ \displaystyle \large{\sum_{k = 1}^{n}ma_k =m\sum_{k = 1}^{n} a_k \: \: \: \sf{(m \: \: is \: \: constant})}[/tex]
Therefore:-
[tex] \displaystyle \large{ \sum_{k = 1}^{3} 5 + \sum_{k = 1}^{3}- 6k} \\ \displaystyle \large{ (5 \times 3) - 6\sum_{k = 1}^{3}k} \\ \displaystyle \large{ 15 - 6\sum_{k = 1}^{3}k}[/tex]
Summation Formula
[tex] \displaystyle \large{ \sum_{k = 1}^{n}k = \frac{1}{2} n(n + 1) }[/tex]
Thus:-
[tex] \displaystyle \large{ 15 - 6\sum_{k = 1}^{3}k} \\ \displaystyle \large{ 15 - 6( \frac{1}{2}(3)(3 + 1) } \\ \displaystyle \large{ 15 - 6( \frac{1}{2}(3)(4)) } \\\displaystyle \large{ 15 - 6(3)(2 )} \\\displaystyle \large{ 15 - 6(6)} \\ \displaystyle \large{ 15 -36 = - 21 } [/tex]
