For the A.P $45, 41, 37, . . .$ find $t_{10}$ and $t_{n + 1}$

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Solution

$a = 45, d = - 4$

We know that $t_{n} = a + (n - 1) d$

$t_{10} = 45 + (10 - 1) (- 4)$

$= 45 + (9 \times - 4) = 45 - 36 = 9$

$t_{n + 1} = a + (n + 1 - 1) d = a + n d$

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