# What is the sum of the arithmetic sequence 174, 168, 162 …, if there are 35 terms?

Jun 19, 2018

$\textcolor{\mathmr{and} a n \ge}{{S}_{35} = 2520}$

#### Explanation:

$\text{sum of n terms of an A.S. } = {S}_{n} = \left(\frac{n}{2}\right) \left[2 a + \left(n - 1\right) d\right]$

$a = 174 , n = 35 , d = 168 - 174 = 162 - 168 = - 6$

Substituting for a, d, n in the formula,

${S}_{35} = \left(\frac{35}{2}\right) \cdot \left[\left(2 \cdot 174\right) + \left(35 - 1\right) \cdot \left(- 6\right)\right]$

${S}_{35} = \left(\frac{35}{2}\right) \left[348 - 204\right] = \left(\frac{35}{2}\right) \cdot 144 = 2520$

Jun 19, 2018

color(orange)(S_(35) = (35/2)(a + a_(35)) = 2520

#### Explanation:

${n}^{t h} \text{ term of an Arithmetic Sequence }$a_n = a + (n-1) d#

$a = 174 , {a}_{2} = 168 , {a}_{3} = 162 , d = 166 - 174 = - 6 , n = 35$

${a}_{35} = 174 + \left(35 - 1\right) \cdot \left(- 6\right) = 174 - 204 = - 30$

$\text{Sum of 35 terms } = {S}_{35} = \left(\frac{n}{2}\right) \left(a + {a}_{35}\right)$

${S}_{35} = \left(\frac{35}{2}\right) \left(174 + \left(- 30\right)\right) = 35 \cdot 72 = 2520$