If the sum of n term of an ap is 3 n square minus 2 and find nth term and 10th term also find difference of an ap
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Answer:
Step-by-step explanation:
given sum of n terms = 3n² + 2n
let first term be a₁, second term be a₂ and so on.....
if n = 1
sum = a₁ (since there is no sum when considering only 1st term)
therefore, a₁ = 3(1)² + 2(1) = 3 + 2 = 5
then, if n = 2
S₂ = a₁ + a₂ = 3(2)² + 2(2) = 12 + 4 = 16
since a₁ = 5
∴ a₂ = 16 - 5 = 11
similarly, if n = 3
S₃ = 3(3)² + 2(3) = 27 + 6 = 33
∴ a₃ = 33 - (a₁ + a₂) = 33 - 16 = 17
∴ we gain that common difference = d = 17 - 11 = 11- 5 = 6
∴ let nth term be an
an = a₁ + (n-1)d
( where a₁ = 5 ; d = 6 and n = n)
∴ an = 5 + (n-1)6
= 5 + 6n - 6
= 6n - 1
hope it's helpful
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