How do you find the 10th partial sum of the arithmetic sequence $40, 37, 34, 31,...$ ?

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Answer 1 · 2018-08-05T10:41:40

$10$ th partial sum is $265$

$S : {40 , 37 ,34,31 ...}$

First term is $a_1=40$ , common difference is $d=37-40=-3$

and number of terms $n=10$

Sum of $n$ terms is $S_n = n/2{2 a_1+(n-1)d}$

$:. S_10 = 10/2{2*40+(10-1)* (-3)}$ or

$S_10 = 5(80-27)= 5 * 53= 265$

$10$ th partial sum is $265$ [Ans]

Answer 2 · 2018-08-05T12:59:29

$color(maroon)(S_(10) = 265$

Arithmetic sequence $40, 37, 34, 31, . . .$

First term $a = 40$

Common difference $d = 37 - 40 = 34 - 37 = 31 - 34 = -3$

$n^(th) term = a_n = a + (n-1) * d$

$a_(10) = a + (10-1) * d = 40 + 9 * -3 = 40 - 27 = 13$

Sum of first n terms $S_n =n/2 * (a + a_n)$

$S_(10) = 10/2 * (a + a_(10)) = 5 * (40 + 13) = 265$

How do you find the 10th partial sum of the arithmetic sequence 40, 37, 34, 31,...40, 37, 34, 31,...?