if the angles of an n sided polygon are in A.P and a>=20 and d>=20 then wat is the maximum possible value of n?
a)12
b)14
c)21
d)cant determine

• Sum of Interior angles of a polygon = (n-2) × 180° where n = number of sides
  There will be n angles which are in A.P. Therefore,
  Since we need to find maximum value of n, put minimum value for a and d.i.e., take a=20 and d=20
  Then sum of the angles
  =n/2[2×20+(n−1)20]
  Therefore,
  10n(n+1)=(n−2)180
  n=14.52
  Since number of sides must be a positive integer, maximum value of n = 14
• 8 years agoHelpfull: Yes(7) No(2)
• Sum of Interior angles of a polygon = (n-2) × 180° where n = number of sides
  
  There will be
  n
  n angles which are in A.P. Therefore,
  Sum of the angles =
  n
  2
  [
  2
  a
  +
  (
  n
  −
  1
  )
  d
  ]
  n2[2a+(n−1)d] where
  a
  ≥
  20
  a≥20 and
  d
  ≥
  20
  d≥20
  
  Since we need to find maximum value of n, put minimum value for
  a
  a and
  d
  d.
  i.e., take
  a
  =
  20
  a=20 and
  d
  =
  20
  d=20.
  
  Then sum of the angles
  =
  n
  2
  [
  2
  ×
  20
  +
  (
  n
  −
  1
  )
  20
  ]
  =
  20
  n
  2
  [
  2
  +
  (
  n
  −
  1
  )
  ]
  =
  10
  n
  (
  n
  +
  1
  )
  =n2[2×20+(n−1)20]=20n2[2+(n−1)]=10n(n+1)
  
  Therefore,
  10
  n
  (
  n
  +
  1
  )
  =
  (
  n
  −
  2
  )
  180
  n
  (
  n
  +
  1
  )
  =
  (
  n
  −
  2
  )
  18
  n
  2
  +
  n
  =
  18
  n
  −
  36
  n
  2
  −
  17
  n
  +
  36
  =
  0
  n
  =
  17
  ±
  √
  (
  −
  17
  )
  2
  −
  4
  ×
  1
  ×
  36
  2
  ×
  1
  =
  17
  ±
  √
  289
  −
  144
  2
  =
  17
  ±
  √
  145
  2
  (∵takling maximum value)
  ≈
  14.52
  10n(n+1)=(n−2)180n(n+1)=(n−2)18n2+n=18n−36n2−17n+36=0n=17±(−17)2−4×1×362×1=17±289−1442=17±1452(∵takling maximum value)≈14.52
  
  Since number of sides must be a positive integer, maximum value of n = 14
  
• 8 years agoHelpfull: Yes(1) No(12)