the 10th term of AP 3,7,11,15, is
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EXPLANATION.
10th term of an A.P.
Series = 3, 7, 11, 15 . . . . .
As we know that,
First term = a = 3.
Common difference = d = b - a = c - b.
Common difference = d = 7 - 3 = 11 - 7.
Common difference = d = 4.
As we know that,
General term of an A.P.
⇒ Tₙ = a + (n - 1)d.
⇒ T₁₀ = a + (10 - 1)d.
⇒ T₁₀ = a + 9d.
Put the values in the equation, we get.
⇒ T₁₀ = 3 + 9(4).
⇒ T₁₀ = 3 + 36.
⇒ T₁₀ = 39.
MORE INFORMATION.
Properties of an A.P.
(1) = If a₁, a₂, a₃ . . . . . aₙ are in A.P then,
(a) = a₁ + k, a₂ + k . . . . . aₙ + k are also in A.P.
(b) = a₁ - k, a₂ - k . . . . . aₙ - k are also in A.P.
(c) = ka₁, ka₂ . . . . . kaₙ are also in A.P.
(d) = a₁/k, a₂/k . . . . . aₙ/k, k ≠ 0 are also in A.P.
(2) = If a₁, a₂ . . . . . and b₁, b₂ . . . . . are two A.P.s then,
(a) = a₁ + b₁, a₂ + b₂, a₃ + b₃ are also in A.P.
(b) = a₁ - b₁, a₂ - b₂, a₃ - b₃ are also in A.P.
(3) = If a₁, a₂, a₃ . . . . . aₙ are in A.P. then,
(a) = a₁ + aₙ = a₂ + aₙ₋₁ = a₃ + aₙ₋₂ = . . . . .
(b) = a(r) = [a_{r - k} + a_{r + k}]/2, 0 ≤ k ≤ n - r.
(c) = If number of terms of any A.P. is odd then sum of the terms is equal to product of middle term and number of terms.
(d) = If number of terms of any A.P. is odd then its middle term is A.M. of first and last term.