The sum of n terms of an ap is 3n2 + ÿn.Find the nth term of an ap
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Sn = 3n^2 + 6n
=> n/2 [ 2a + (n-1) d] = n (3n + 6)
=> [ 2a + (n-1) d] = 2 (3n +6)
=> 2a + (n-1) d = 6n + 12
=> 2a + (n-1)d = 18 + 6n - 6
=> 2a + (n-1)d = 18 + 6(n-1)
On Comparing both sides, we get
2a = 18
=> a = 9
And, d = 6
Tn = a + (n-1) d
= 9 + (n-1) 6
= 9 + 6n - 6
= 3 + 6n
Hope it helps you
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=> n/2 [ 2a + (n-1) d] = n (3n + 6)
=> [ 2a + (n-1) d] = 2 (3n +6)
=> 2a + (n-1) d = 6n + 12
=> 2a + (n-1)d = 18 + 6n - 6
=> 2a + (n-1)d = 18 + 6(n-1)
On Comparing both sides, we get
2a = 18
=> a = 9
And, d = 6
Tn = a + (n-1) d
= 9 + (n-1) 6
= 9 + 6n - 6
= 3 + 6n
Hope it helps you
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