8th term of an a.p is 37 and 15th term is 15 more than the 12th term . find the A.p hence find the sum of the A.p
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First condition,
T8 = 37
=> a + 7d = 37 ------(1)
Second condition,
T15 - T12 = 15
=> a +14d - (a+11d) = 15
=> a + 14d - a - 11d = 15
=> 3d = 15
=> d = 5
On putting the value of d in equation 1, we get
a = 2
Required AP is 2,7,12,.........
Sn = n/2 [ 2*2 + (n-1)5]
= n/2 ( 4 +5n - 5)
= n/2 ( 5n - 1)
= (5n^2 - n) /2
T8 = 37
=> a + 7d = 37 ------(1)
Second condition,
T15 - T12 = 15
=> a +14d - (a+11d) = 15
=> a + 14d - a - 11d = 15
=> 3d = 15
=> d = 5
On putting the value of d in equation 1, we get
a = 2
Required AP is 2,7,12,.........
Sn = n/2 [ 2*2 + (n-1)5]
= n/2 ( 4 +5n - 5)
= n/2 ( 5n - 1)
= (5n^2 - n) /2
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