Question 18 of 51beginner🔧 ApplyNumerical3 marks

Using the formula tn=a+(n−1)×dt_n = a + (n - 1) \times d, find the nthn^{th} term of the following arithmetic progressions. (i) 12,52,92,132,…\frac{1}{2}, \frac{5}{2}, \frac{9}{2}, \frac{13}{2}, \ldots (ii) 1.5, 3.5, 5.5, 7.5, …

(i)
Answer

(i) First term (a) = 1/2 Common difference (d) = 5/2 - 1/2 = 2 Using the formula t_n = a + (n - 1) × d: t_n = 1/2 + (n - 1) × 2 t_n = 1/2 + 2n - 2 t_n = 2n - 3/2

So, the nth term is 2n - 3/2.

(ii)
Answer

(ii) First term (a) = 1.5 Common difference (d) = 3.5 - 1.5 = 2 Using the formula t_n = a + (n - 1) × d: t_n = 1.5 + (n - 1) × 2 t_n = 1.5 + 2n - 2 t_n = 2n - 0.5

So, the nth term is 2n - 0.5.

Explanation

The question requires calculating the nth term of two arithmetic progressions using the formula t_n = a + (n - 1) × d. For part (i), the first term a is 1/2 and the common difference d is 2. Substituting these into the formula gives t_n = 2n - 3/2. For part (ii), the first term a is 1.5 and the common difference d is 2. Substituting these gives t_n = 2n - 0.5.

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