Find the $10^{th}$ and $26^{th}$ terms of the A.P.: $3, 8, 13, 18, \dots$                                                                                   [Page No. 185]

First term $a = 3$, common difference $d = 5$

$n^{\text{th}}$ term: $a_n = a + (n - 1)d$

The $10^{\mathrm{th}}$ term: $a_{10} = 3 + (10 - 1) \times 5$

$$= 3 + 45 = 48$$

The $26^{\text{th}}$ term: $a_{26} = 3 + (26 - 1) \times 5$

$$= 3 + 125 = 128$$

Caution
Students should be careful with the nth-term formulas: for an AP, $a
eq a + (n - 1)d$, and for a GP,
  $T = ar
eq$Using$n$in place of$(n - 1)$ is a common error.