An algebra problem by Paola Ramírez

Algebra Level 2

The 1 st 1^\text{st} term and the 4 th 4^\text{th} terms of an arithmetic progression are 5 and 15 respectively. Find the value of the 1 0 th 10^\text{th} term.


The answer is 35.

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3 solutions

Ashish Menon
Jun 6, 2016

Let a a be the first term and d d be the common difference.
a = 5 1 a 4 = a + 3 d = 15 2 Substituting 1 in 2 : 5 + 3 d = 15 3 d = 10 d = 10 3 a 10 = a + 9 d = 5 + 9 × 10 3 = 5 + 30 = 35 \begin{aligned} a & = 5 \longrightarrow \boxed{1}\\ \\ a_4 = a + 3d & = 15 \longrightarrow \boxed{2}\\ \\ \text{Substituting} \ \boxed{1} \ \text{in} \ \boxed{2}:-\\ 5 + 3d & = 15\\ 3d & = 10\\ d & = \dfrac{10}{3}\\ \\ \implies a_{10} & = a + 9d\\ & = 5 + 9×\dfrac{10}{3}\\ & = 5 + 30\\ & = \color{#3D99F6}{\boxed{35}} \end{aligned}

Typo: 5 + 30 5+30

Hung Woei Neoh - 5 years ago

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Thnks! I corrected it.

Ashish Menon - 5 years ago

Given: a 1 = 5 a_1=5 and a 4 = 15 a_4=15

Formula: a n = a m + ( n m ) d a_n=a_m+(n-m)d

Substitute: 15 = 5 + ( 4 1 ) d 15=5+(4-1)d

d = 10 3 d=\dfrac{10}{3}

Finally.

a 10 = a 1 + 9 d = 5 + 9 ( 10 3 ) = a_{10}=a_1+9d=5+9\left(\dfrac{10}{3}\right)= 35 \color{#D61F06}\boxed{35}

A former brilliant member? This is my solution, I deleted my account and created a new one.

A Former Brilliant Member - 1 year, 4 months ago
Hung Woei Neoh
Jun 5, 2016

a 4 = a 1 + 3 d 15 = 5 + 3 d 3 d = 10 a 10 = a 1 + 9 d = a 1 + 3 ( 3 d ) = 5 + 3 ( 10 ) = 35 a_4 = a_1 + 3d\\ 15=5+3d\\ 3d=10\\ a_{10} = a_1 + 9d\\ =a_1+ 3(3d)\\ =5+3(10)\\ =\boxed{35}

Perfect solution(+1)

Ashish Menon - 5 years ago

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