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Algebra Level 2

Given that the $5^\text{th}$ and the $12^\text{th}$ of an arithmetic progression are 30 and 65 respectively, what is the sum of the first 20 terms?

1000 1755 1620 1225 1150

3 solutions

A Former Brilliant Member
May 25, 2016

Using A.P.

$a_5=a+4d=30$
$a_{12}=a+11d=65$

Solving we get,
$a=10$
$d=5$

$S_{20}=\dfrac{20[(10×2)+(20-1)×5]}{2}=200+950=\boxed{1150}$

You missed out a bracket.

It should be: $S_{20} = \dfrac{20\color{#3D99F6}{(}(10 \times 2) + (20 -1) \times 5 \color{#3D99F6}{)}}{2}$

Hung Woei Neoh - 5 years ago

Done.... :)

A Former Brilliant Member - 5 years ago

A Former Brilliant Member
Apr 2, 2018

$a_n=a_m+(n-m)d$

$65=30+(12-5)d$

$d=5$

$a_1=30+(1-4)(5)$

$a_1=10$

$s=\dfrac{20}{2}[2(10)+(20-1)(5)]=\boxed{1150}$

Ma Pm
May 26, 2016

(65-30)/7=5;

d=5;

a1=a5-(n)x(d)=30-4x5=10;

a20=a5+(m)x(d)=30+15x5=105;

S=((10+105)x20)/2=1150;