# Find:

Question:

Find:

(i) the 20 th term of the AP $9,13,17,21, \ldots .$

(ii) the 35 th term of the AP $20,17,14,11$,

(iii) the 18 th term of the AP $\sqrt{2}, \sqrt{18}, \sqrt{50}, \sqrt{98}, \ldots$.

(iv) the 9 th term of the AP $\frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \frac{9}{4}, \ldots .$

(v) the 15 th term of the AP $-40,-15,10,35, \ldots .$

Solution:

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

First term, a = 9

Common difference, d = 13 − 9 = 4

nth term of the AP, an = a + (− 1)d = 9 + (− 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

nth term of the AP, an = a + (− 1)d = 20 + (− 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 $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, 7 \sqrt{2}, \ldots .$

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

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

$n^{\text {th }}$ term of the $\mathrm{AP}, a_{n}=a+(n-1) d=\sqrt{2}+(n-1) \times 2 \sqrt{2}$

$\therefore$ 18th term of the AP, $a_{18}=\sqrt{2}+(18-1) \times 2 \sqrt{2}=\sqrt{2}+34 \sqrt{2}=35 \sqrt{2}=\sqrt{2450}$

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

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

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

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

$\therefore$ 9th term of the AP, $a_{9}=\frac{3}{4}+(9-1) \times \frac{1}{2}=\frac{3}{4}+4=\frac{19}{4}$

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

First term, a = −40

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

nth term of the AP, an = a + (− 1)d = −40 + (− 1) × 25

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