Find the 30th term of the AP: $10, 7, 4, . . . . .$

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Solution

Given $A P : - 10, 7, 4....$
Let $a_{n}$ be the $n^{t h}$ term of the given $A P$ $d$ be the common difference
$a_{1} = 10, a_{2} = 7, a_{3} = 4$
$a_{n} = a_{1} + (n - 1) d$
$a_{2} = a_{1} + (2 - 1) d$
$d = a_{2} - a_{1} = 7 - 10 = - 3$
$d = - 3$
$a_{30} = a_{1} + (30 - 1) d$
$= 10 + (29) (- 3)$
$= 10 - 87$
$= 77$
Ans :- $30 t h$ term of the given $A P$ is $77$

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