Question

Find:

(i) the 20th term of the AP 9, 13, 17, 21, ... .
(ii) the 35th term of the AP 20, 17, 14, 11, ... .
(iii) the 18th term of the AP $\sqrt{2},\sqrt{18},\sqrt{50},\sqrt{98}$, ... .
(iv) the 9th term of the AP $\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4}$, ... .
(v) the 15th term of the AP −40, −15, 10, 35, ... .

Answer 1

(i) The given AP is 9, 13, 17, 21, ... .

First term, a = 9

Common difference, d = 13 − 9 = 4

n^th term of the AP, a_n = a + (n − 1)d = 9 + (n − 1) × 4

∴ 20th term of the AP, a₂₀ = 9 + (20 − 1) × 4 = 9 + 76 = 85

(ii) The given AP is 20, 17, 14, 11, ... .

First term, a = 20

Common difference, d = 17 − 20 = −3

n^th term of the AP, a_n = a + (n − 1)d = 20 + (n − 1) × (−3)

∴ 35th term of the AP, a₃₅ = 20 + (35 − 1) × (−3) = 20 − 102 = −82

(iii) The given AP is $\sqrt{2},\sqrt{18},\sqrt{50},\sqrt{98}$, ... .

This can be re-written as $\sqrt{2},3\sqrt{2},5\sqrt{2},7\sqrt{2}$, ... .

First term, a = $\sqrt{2}$

Common difference, d = $3\sqrt{2}-\sqrt{2}=2\sqrt{2}$

n^th term of the AP, a_n = a + (n − 1)d = $\sqrt{2}+\left(n-1\right)\times 2\sqrt{2}$

∴ 18th term of the AP, a₁₈ = $\sqrt{2}+\left(18-1\right)\times 2\sqrt{2}=\sqrt{2}+34\sqrt{2}=35\sqrt{2}=\sqrt{2450}$

(iv) The given AP is $\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4}$, ... .

First term, a = $\frac{3}{4}$

Common difference, d = $\frac{5}{4}-\frac{3}{4}=\frac{2}{4}=\frac{1}{2}$

n^th term of the AP, a_n = a + (n − 1)d = $\frac{3}{4}+\left(n-1\right)\times \left(\frac{1}{2}\right)$

∴ 9th term of the AP, a₉ = $\frac{3}{4}+\left(9-1\right)\times \frac{1}{2}=\frac{3}{4}+4=\frac{19}{4}$

(v) The given AP is −40, −15, 10, 35, ... .

First term, a = −40

Common difference, d = −15 − (−40) = 25

n^th term of the AP, a_n = a + (n − 1)d = −40 + (n − 1) × 25

∴ 15th term of the AP, a₁₅ = −40 + (15 − 1) × 25 = −40 + 350 = 310