If the sum of the first nterms of an AP is 3 + 4n2.Find the 3rd, the 10th andthe nth terms.

Sum of first n terms of an AP – Sn = 3 + 4n²

the 3rd, the 10th andthe nth terms.

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.

step-by-step explanation:

21.4 i did it already for s1

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