Question

# Which term of the AP: 121,117,113,..., is its first negative term. Find $n$ for ${a_n} =0$

Hint: An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. Use the formula of the general term to find the value of n in this question.
Using the formula of general term in this question, ${T_n} = a + \left( {n - 1} \right)d = 0$
We get,
${T_n} = 121 + \left( {n - 1} \right)\left( { - 4} \right) = 0$
Therefore, on solving this becomes,
$n = \dfrac{{121}}{4} + 1$
On further solving,
We get,
$n = 31.25$

Note: Since 31.25th term does not exist, make sure to round off the number to the nearest number.
This clearly means the value of can be taken roughly as $32$,
Therefore,
$n \approx 32$