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

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Answer 1 · 2018-06-19T08:48:12

$S_{35} = 2520$ $color(orange)(S_(35) = 2520)$

$sum of n terms of an A.S. = S_{n} = (\frac{n}{2}) [2 a + (n - 1) d]$ $"sum of n terms of an A.S. " = S_n = (n/2) [2a + (n-1)d]$

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

Substituting for a, d, n in the formula,

$S_{35} = (\frac{35}{2}) \cdot [(2 \cdot 174) + (35 - 1) \cdot (- 6)]$ $S_(35) = (35/2) * [(2* 174) + (35-1)*(-6)]$

$S_{35} = (\frac{35}{2}) [348 - 204] = (\frac{35}{2}) \cdot 144 = 2520$ $S_(35) = (35/2) [348 - 204] = (35/2) * 144 = 2520$

Answer 2 · 2018-06-19T08:53:58

$S_{35} = (\frac{35}{2}) (a + a_{35}) = 2520$ $color(orange)(S_(35) = (35/2)(a + a_(35)) = 2520$

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

$a = 174, a_{2} = 168, a_{3} = 162, d = 166 - 174 = - 6, n = 35$ $a = 174, a_2 = 168, a_3 = 162, d = 166 - 174 = -6, n = 35$

$a_{35} = 174 + (35 - 1) \cdot (- 6) = 174 - 204 = - 30$ $a_(35) = 174 + (35-1) * (-6) = 174 - 204 = - 30$

$Sum of 35 terms = S_{35} = (\frac{n}{2}) (a + a_{35})$ $"Sum of 35 terms " = S_(35) = (n/2)(a + a_(35))$

$S_{35} = (\frac{35}{2}) (174 + (- 30)) = 35 \cdot 72 = 2520$ $S_(35) = (35/2)(174 + (-30)) = 35 * 72 = 2520$