Question

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

Answer 1

5th term is -25

and sum of first three terms is -21

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 * 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 * 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 = (1)/(2) n(n + 1) }`

Thus:-

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