Find the number of terms of this progression

Algebra Level pending

The sum of an arithmetic progression is 1645 -1645 . If the 2 n d 2^{nd} and 1 0 t h 10^{th} terms are 35 -35 and 155 -155 , respectively, find the number of terms of this progression.


The answer is 14.

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2 solutions

Relevant wiki: Arithmetic Progressions

Using the formula a n = a m + ( n m ) d a_n=a_m+(n-m)d , we have

155 = 35 + ( 10 2 ) d -155=-35+(10-2)d \color{#D61F06}\implies d = 15 d=-15

Using again the formula, we have

a 1 = 35 + ( 1 2 ) ( 15 ) = 20 a_1=-35+(1-2)(-15)=-20

Using the formula s = n 2 [ 2 a 1 + ( n 1 ) d ] s=\dfrac{n}{2}[2a_1+(n-1)d] , we have

1645 = n 2 [ 2 ( 20 ) + ( n 1 ) ( 15 ) ] -1645=\dfrac{n}{2}[2(-20)+(n-1)(-15)]

Simplifying this will end in a quadratic equation

15 n 2 + 25 n 3290 = 0 15n^2+25n-3290=0

Using any method to solve for n n , we find that n = 14 n=\boxed{14} .

Razing Thunder
Jul 2, 2020

final equation (x)/(2)(-40+(x-1)-15)=-1645

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