Question

IN an AP
given a = 2, d= 8, Sn = 90, find n and an
​

Answer 1

Answer:-

N= 5 and an = 34

Step-by-step explanation:

Given:-

• In an AP, a = 2, d=8, Sn=90
• To find:- n and an

Solⁿ:-

Since we know that Sn = n/2+(2a+(n-1)d)

=> Sn=90

Therefore, 90= n/2(2a+(n-1)d)

=> 180 = n(2×2+(n-1)8)

=> 180 = n(4+8n-8)

=> 180 = n(8n-4)

=> 180 = 8n²-4n

=> Bringing/Transposing 180 to RHS

=> 8n²-4n-180 = 0

=> Dividing throughout by 4

=> 2n²-n-45 = 0

=> Using middle term split

=> 2n²-10n+9n-45 = 0

=> 2n(n-5)+9(n-5) = 0

=>(2n+9)(n-5) = 0

=> n = -9/2[not possible] and 5

Therefore, n=5

Since n = 5 So, an = a+(n-1)d

=> a5 = 2+(5-1)8

=> a5 = 2+(4×8)

=> a5 = 2+32 = 34

therefore, an/a5 = 34 and n = 5

I Hope It Helps You!!!!.... :):)

Answer 2

Given

⇒First term(a) = 2

⇒Common Difference (d) = 8

⇒Sum of nth term (Sₙ) = 90

To Find

⇒Number of term (n) , Tₙ

Formula which we use in this question

⇒Sₙ = n/2{2a+(n-1)d}

⇒Tₙ = a + (n-1)d

Now we have to find 'n' So take

⇒Sₙ = n/2{2a+(n-1)d}

put the value on formula

⇒90=n/2{2×2 +(n-1)×8}

⇒90×2 = n{4+8n-8}

⇒180 = n{8n-4}

⇒180=8n² - 4n

⇒8n² - 4n - 180 = 0

⇒4(2n²-n-45)=0

⇒2n²-n-45=0

Now factorise the quadratic equation

⇒2n²-10n+9n-45=0

⇒2n(n-5)+9(n-5)=0

⇒(2n+9)(n-5)=0

⇒n=5 and n=-9/2

So we take n = 5 because 'n' is never in negative and fraction

Now we have to find Tₙ

⇒Tₙ = a + (n-1)d

Where

⇒a= 2 , d = 8 and n = 5

put the value on formula

⇒T₅ = 2+(5-1)8

⇒T₅ = 2+4(8)

⇒T₅=2+32

⇒T₅ = 34