In the arithmetic progression series $$ \{a_n\} $$ , if $$ a_1=\dfrac{1}{3} $$ , $$ a_2+a_5=4 $$ , $$ a_n=33 $$ , what is $$ n $$ ?

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In the arithmetic progression series $$ \{a_n\} $$ , if $$ a_1=\dfrac{1}{3} $$ , $$ a_2+a_5=4 $$ , $$ a_n=33 $$ , what is  $$ n $$ ?

Uniteasy Admin · Accepted Answer

50

$$ a_1=\dfrac{1}{3} $$ , $$ a_2+a_5=4 $$

$$ a_2 = a_1+(2-1)d=a_1+d $$n
$$ a_5 = a_1+(5-1)d=a_1+4d $$n

Subtract (1) from (2) 
 
$$ a_5-a_2 = 3d $$n
 $$ a_2+a_5=4 $$n

Subtract (3) from (4) 

$$ a_2 =  \dfrac{4-3d}{2}  $$n

subtact (1) from (5) 

$$ \dfrac{4-3d}{2}=a_1+d $$
$$ \dfrac{4-3d}{2}=\dfrac{1}{3}  +d $$

$$ 12-9d=2+6d $$
$$ d = \dfrac{2}{3}  $$

$$ a_n = a_1 +(n-1)d = 33 $$
$$ = \dfrac{1}{3} + (n-1)\dfrac{2}{3}  = 33 $$

$$ 1+2n-2 = 99 $$

$$n = 50 $$

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