the sum of the first n terms of an ap is (3n×n÷2+5n÷2. find the nthand the 25th term
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Sum of n terms of an A.P = n [2a+ (n-1)d] / 2 = an + nd(n-1)/2
= an + n²d/2 - nd/2 ⇒ n(a-d/2) + n²d/2
Given , Sum = 3n²/2 + 5n/2
Comparing with actual expression
fetches d = 3 and a - d/2 = 5/2 ⇒ a- 3/2 = 5/2 ⇒ a = 4
a₂₅ = a + (25-1)d = 4 + 24x3 = 76
an = a + (n-1)d = 4 + (n-1)3
Hope this helps.
= an + n²d/2 - nd/2 ⇒ n(a-d/2) + n²d/2
Given , Sum = 3n²/2 + 5n/2
Comparing with actual expression
fetches d = 3 and a - d/2 = 5/2 ⇒ a- 3/2 = 5/2 ⇒ a = 4
a₂₅ = a + (25-1)d = 4 + 24x3 = 76
an = a + (n-1)d = 4 + (n-1)3
Hope this helps.
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