If the nth term of a progression is (4n − 10), show that it is an AP. Find its
(i) first term, (ii) common difference, and (iii) 16th term.

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Solution

Tn = (4n - 10) [Given]
T₁ = (4 ⨯ 1 - 10) = -6
T₂ = (4 ⨯ 2 - 10) = -2
T₃ = (4 ⨯ 3 - 10) = 2
T₄ = (4 ⨯ 4 - 10) = 6
Clearly, [ -2 - (-6)] = [2 - (-2)] = [6 - 2] = 4 (Constant)

So, the terms -6, -2, 2, 6,... forms an AP.

Thus, we have;
(i) First term = -6
(ii) Common difference = 4
(iii) T₁₆ = a + (n -1)d = a + 15d = - 6 + 15 ⨯ 4 = 54

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If the nth term of a progression is (4n − 10), show that it is an AP. Find its (i) first term, (ii) common difference, and (iii) 16th term.

If the nth term of a progression is (4n − 10), show that it is an AP. Find its
(i) first term, (ii) common difference, and (iii) 16th term.