Given:
Sum of first n terms of an AP – Sn = 3 + 4n²
Find:
the 3rd, the 10th andthe nth terms.
Solution:
Substitute the value of n as 1 to find the first term of the given AP.
(Sum of first 1 terms is nothing but the first term).
Hence,
⟹ S₁ = a = 3 + (4)(1)²
⟹ a = 3 + 4
⟹ a = 7
Now,
Substitute the value of n as 2 to find the sum of first 2 terms.
⟹ S₂ = a + a₂ = 3 + 4(2)²
⟹ a + a₂ = 3 + 16
Substitute the value of a here.
⟹ 7 + a₂ = 19
⟹ a₂ = 19 - 7
⟹ a₂ = 12
We know,
nth term of an AP – an = a + (n - 1)d
So,
a₂ = a + (2 - 1)d = a₂ = a + d.
Subtract a from a₂.
⟹ a + d - a = 12 - 7
⟹ d = 5
Now,
nth term = an = 12 + (n - 1)(5)
⟹ an = 12 + 5n - 5
⟹ an = 5n + 7
Substitute n as 3 & 10 , to find the 3rd & 10th terms.
★ a₃ = 5(3) + 7
⟹ a₃ = 8 + 7
⟹ a₃ = 15
★ a₁₀ = 5(10) + 7
⟹ a₁₀ = 50 + 7
⟹ a₁₀ = 57
Therefore,
Third term is 15
Tenth term is 57
nth term is 5n + 7.
I hope it will help you.
Regards.