Question

solve this
In an AP

Answer 1

Hi...

Here is your answer... 

Given that ,

a = 2 , d = 8 , Sn = 90 

n= ? and $a_{n} =?$

$S_{n} =90 \\ \\ \frac{n}{2} [2a +(n-1)d]=90\\ \\ {n} [2(2) +(n-1)8]=90\times 2 \\ \\ {n} ( 4 +8n-8 ) = 180 \\ \\ n(8n - 4)=180 \\ \\8 n^{2} -4n -180 = 0 \\ \\ \implies 2n^{2} -n -45 = 0 \\ \\ \implies 2n^{2} -n -45 = 0\\ \\ \implies 2n^{2} -10n + 9n -45 = 0 \\ \\ \implies 2n(n-5)+9(n-5)=0 \\ \\ \implies (2n+9)(n-5)=0 \\ \\ \implies 2n+9=0 \:\:\: or \:\:\:n-5=0 \\ \\ \implies n= \frac{-9}{2} \:\:\:or \:\:\: n=5 \\ \\ \implies n=5 \:\:\:\:\: [\: \because n\: \:cannot \:\:be\:\: negative\:]$

Now,

$a_{n} = a + (n-1)d \\ \\ a_{n} = 2 + (5-1)8 \\ \\a_{n} = 2 + 4(8) \\ \\ a_{n} = 2+32 \\ \\ a_{n} = 34 \\ \\ Hence,\\ \\ n= 5 \:,\: a_{n} = 34$