An algebra problem by Paola Ramírez

Algebra Level 2

The 3 rd 3^\text{rd} and 6 th 6^\text{th} terms of an arithmetic progression are 94 and 85 respectively. Find the value of the 1 5 th 15^\text{th} term.


The answer is 58.

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4 solutions

Ashish Menon
Jun 6, 2016

Let a a be the first term and d d be the common difference.
a 3 = a + 2 d = 94 1 a 6 = a + 5 d = 85 2 Subtracting 2 from 1 : 3 d = 9 d = 3 Substituting d = -3 in 1 : a + 2 ( 3 ) = 94 a 6 = 94 a = 100 a 15 = a + 14 d = 100 + 14 ( 3 ) = 100 42 = 58 \begin{aligned} a_3 = a + 2d & = 94 \longrightarrow \boxed{1}\\ \\ a_6 = a + 5d & = 85 \longrightarrow \boxed{2}\\ \\ \text{Subtracting} \ \boxed{2} \ \text{from} \ \boxed{1}:-\\ -3d & = 9\\ d = -3\\ \\ \text{Substituting d = -3 in} \ \boxed{1}:-\\ a + 2(-3) & = 94\\ a - 6 & = 94\\ a & = 100\\ \\ \implies a_{15} & = a + 14d\\ & = 100 + 14(-3)\\ & = 100 - 42\\ & = \color{#3D99F6}{\boxed{58}} \end{aligned}

Rohit Ner
Jun 5, 2016

a 15 = a 2 + 12 ( a 6 a 3 3 ) = 94 + 12 ( 3 ) = 58 \begin{aligned} a_{15}&=a_2+12\left(\dfrac{a_6-a_3}{3}\right)\\&=94+12(-3)\\&\huge\color{#3D99F6}{=\boxed{58}}\end{aligned}

Given: a 3 = 94 ; a 6 = 85 \large{\text{Given:}~a_3=94; a_6=85}

Formula: a n = a m + ( n m ) d \large{\text{Formula:}~a_n=a_m+(n-m)d}

We have \large{\text{We have}}

85 = 94 + ( 6 3 ) d \large{85=94+(6-3)d}

d = 3 \large{d=-3}

a 15 = a 3 + 12 d = 94 + 12 ( 3 ) = \large{a_{15}=a_3+12d=94+12(-3)=} 58 \color{#D61F06}\huge{\boxed{58}}

Good solution. There is no need of involving the first term then having two equations and two unknowns. You immediately calculated the value of d d .

A Former Brilliant Member - 3 years, 7 months ago
Hung Woei Neoh
Jun 6, 2016

a 3 = 94 a + 2 d = 94 a_3 = 94 \implies a+2d = 94\implies Eq.(1)

a 6 = 85 a + 5 d = 85 a_6 = 85 \implies a+5d = 85 \implies Eq.(2)

Eq.(2) - Eq.(1):

a + 5 d ( a + 2 d ) = 85 94 3 d = 9 a+5d - (a+2d) = 85 - 94\\ 3d = -9

a 15 = a + 14 d = a + 2 d + 12 d = a 3 + 4 ( 3 d ) = 94 + 4 ( 9 ) = 94 36 = 58 a_{15} = a + 14d\\ =a+2d+12d\\ =a_3 + 4(3d)\\ =94+4(-9)\\ =94-36\\ =\boxed{58}

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