If the sum of n terms of an AP is 3n^2+ 5n then which of its term is 164
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Answer:-
Given:
Sum of first n terms of an AP = 3n² + 5n
We know that,
Sum of first n terms of an AP = n/2 * [ a + l ]
Where,
- a = first term
- l = nth term or last term.
Substitute the value of n as 1 to find the sum of first 1 terms (first term)
⟶ S₁ = a = 3(1)² + 5(1)
⟶ a = 3 + 5
⟶ a = 8
Hence,
⟶ n/2 * [ 8 + l ] = 3n² + 5n
⟶ 8 + l = n(3n + 5) * (2/n)
⟶ 8 + l = 6n + 10
⟶ l = 6n + 10 - 8
⟶ l = 6n + 2
Now,
We have to find which term is 164.
We have: l = 6n + 2
So,
⟶ 6n + 2 = 164
⟶ 6n = 164 - 2
⟶ 6n = 162
⟶ n = 162/6
⟶ n = 27
Therefore, 164 is the 27th term of the given AP.
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