In an A.P. :
Given a = 2, d = 8, Sn = 90, find ‘n’ and an
Here,   a = 2, d = 8, Sn = 90
We know that,
      Sn =  [2 x 2 + (n -1)d]
   180  =n[2 x 2 (n - 1) x 8]
   180 = n(8n - 4)
   180 = 8n2 - 4n
   8n2 - 4n - 180 = 0
   2n2 - n - 45 = 0   [Dividing by 4]
  2n(n - 5) (2n + 9) = 0
  n - 5 = 0  or 2n + 9 = 0
  n = 5 or n =Â
But number of terms cannot be (- ve)
Now, Â a5Â = a+ 4d
       = 2 + 4 + 8
Hence, Â Â n = 5 Â Â Â
an = 34
In an A.P.Â
Given a = 8, an = 62, Sn = 210, find ‘n’ and ‘d’.
Here a = 8
Sn = 210
We know that, an = a + (n - 1)d
  62 - 8 = (n - 1)d
  54 = (n -d)d          ...(i)
And,  Sn =  '[ 2a - (n - 1) d
 210 = [ 2 x 8 + (n - 1) d]
 420 = n[16 +(n - 1)d] .......(ii)
From (i) and (ii), we get
420 Â = n[ 16 + 54]
  420  = 70n
Putting this value of n in (i) we get
54 = (6 -1) x d
   5d = 54
   d =Â
Hence, Â n = 6 and d =Â
In an A.P.Â
Given an = 4, d = 2, Sn = –14, find ‘n’ and ‘a’.
Here,  an = 4,   d = 2
Sn = –14
We know that, Â
  an = a + (n -1)d
 4 = a + (n -1) x 2
 4 = a + 2n - 2
 a + 2n = 6  a = 6 - 2 n   ....(i)
And. Â Â Sn =Â Â [2a+ (n - 1)d]
      - 14 =Â
      - 28 = n[2a + 2 (n -1) ]
      - 28 = 2n [a + (n -1)]
      - 14 = n[a + (n - 1)]    ......(ii)
Putting the value of (i) in (ii), we get,
– 14 = n [6 – 2n + n – 1]
⇒ – 14 = n[5 – n]
⇒ –14 = 5n – n2
⇒ n2 –5n – 14 = 0
⇒ (n – 7) (n + 2) = 0
⇒ n – 7 = 0 or n + 2 = 0
⇒ n = 7 or n = –2
But n cannot be –ve. ∴ n = 7.
Putting this value of V in equation (i), we get
a + 2 x 7 = 6
⇒ a = 6 – 14 = – 8
Hence, n = 7 and a = –8
In an A.P.Â
Given a = 3, n = 8, S = 192, find ‘d’.
Here a = 3, n = 8
S = 192
We know that, Â
   Â
  6 +7d = 48
Hence, d = 6 Ans.
In an A.P.Â
Given l = 28, S = 144, n = 9, find ‘a’.
Here, l = 28, S = 144
n = 9
l = a + (n – 1)d
28 = a + (9 – 1)d
⇒ 28 = a + 8d ...(i)
And,
   Â
  32 = 2(a + 4d)
  a + 4d = 16                  .....(ii)
Solving (i) and (ii) we get,
   a = = 32 -28 = 4
Hence ,   a  =  4