Question

An=5-6n fifth term and sum of first three terms

Answer 1

Answer:

5th term is -25

and sum of first three terms is -21

Step-by-step explanation:

We are given the sequence:

$\displaystyle \large{a_n = 5 - 6n}$

To find 5th term, substitute n = 5.

$\displaystyle \large{a_5 = 5 - 6(5)} \\ \displaystyle \large{a_5 = 5 - 30} \\ \displaystyle \large{a_5 = - 25}$

Therefore, fifth term is 25.

Next, to find the sum of first three terms, we will introduce sigma.

$\displaystyle \large{a_1 + a_2 + a_3 + ... + a_n = \sum_{k = 1}^{n} a_k}$

Our ak is 5-6k

Since we want to find sum of first three terms:-

$\displaystyle \large{ \sum_{k = 1}^{3}( 5 - 6k)}$

Expand Sigma in.

$\displaystyle \large{ \sum_{k = 1}^{3} 5 + \sum_{k = 1}^{3}- 6k}$

Property of Summation

$\displaystyle \large{ \sum_{k = 1}^{n} m = m \times n \: \: \: \sf{(m \: \: is \: \: constant})} \\ \displaystyle \large{\sum_{k = 1}^{n}(a_k + b_k) = \sum_{k = 1}^{n}a_k + \sum_{k = 1}^{n}b_k} \\ \displaystyle \large{\sum_{k = 1}^{n}ma_k =m\sum_{k = 1}^{n} a_k \: \: \: \sf{(m \: \: is \: \: constant})}$

Therefore:-

$\displaystyle \large{ \sum_{k = 1}^{3} 5 + \sum_{k = 1}^{3}- 6k} \\ \displaystyle \large{ (5 \times 3) - 6\sum_{k = 1}^{3}k} \\ \displaystyle \large{ 15 - 6\sum_{k = 1}^{3}k}$

Summation Formula

$\displaystyle \large{ \sum_{k = 1}^{n}k = \frac{1}{2} n(n + 1) }$

Thus:-

$\displaystyle \large{ 15 - 6\sum_{k = 1}^{3}k} \\ \displaystyle \large{ 15 - 6( \frac{1}{2}(3)(3 + 1) } \\ \displaystyle \large{ 15 - 6( \frac{1}{2}(3)(4)) } \\\displaystyle \large{ 15 - 6(3)(2 )} \\\displaystyle \large{ 15 - 6(6)} \\ \displaystyle \large{ 15 -36 = - 21 }$