Question

Which term of the AP: 121, 117, 113, … is its first negative term?

Answer 1

Here, we have to find which term of the given A.P. is its first negative term. Let the ${p^{{\text{th}}}}$ term of an A.P. be its first negative term. First, we will find the common difference of the A.P. Then, using the formula for ${n^{{\text{th}}}}$ term of an A.P., we will find the ${p^{{\text{th}}}}$ term of the A.P. Since the term is the first negative term, we can form an inequation in terms of $p$. We will solve this inequation to get the smallest value of $p$ which is a natural number.

Formula Used: The ${n^{{\text{th}}}}$ term of an A.P. is given by the formula ${a_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term of the A.P. and $d$ is the common difference.

Complete step-by-step answer:
Let the ${p^{{\text{th}}}}$ term of an A.P. be its first negative term.
First, we will find the common difference of the A.P.
We know that the difference between any two consecutive terms of the AP is the common difference.
Therefore, we get
Common difference $=$ Second term $-$ First term
Substituting 121 as the first term and 117 as the second term, we get
Common difference $= 117 - 121 = - 4$
Now, the ${n^{{\text{th}}}}$ term of an A.P. is given by the formula ${a_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term of the A.P. and $d$ is the common difference.
We will use this formula to find the ${p^{{\text{th}}}}$ term of an A.P.
Substituting $n = p$, $a = 121$, and $d = - 4$ in the formula for ${n^{{\text{th}}}}$ term of an A.P., we get
$\Rightarrow {a_p} = 121 + \left( {p - 1} \right)\left( { - 4} \right)$
Multiplying the terms of the expression using the distributive law of multiplication, we get
$\Rightarrow {a_p} = 121 - 4p + 4$
Adding 121 and 4, we get
$\Rightarrow {a_p} = 125 - 4p$
The ${p^{{\text{th}}}}$ term of an A.P. be its first negative term.
Therefore, we get
$\Rightarrow {a_p} < 0$
Substituting ${a_p} = 125 - 4p$ in the inequation, we get
$\Rightarrow 125 - 4p < 0$
Adding $4p$ on both sides of the inequation, we get

$$\Rightarrow 125 < 4p \\\$$

Dividing both sides of the inequation by 4, we get

$$\Rightarrow \dfrac{{125}}{4} < \dfrac{{4p}}{4} \\\ \Rightarrow 31.25 < p \\\$$

Since $p$ has to be a natural number, we get
$\therefore p = 32$
Therefore, the 32nd term of the given A.P. is its first negative term.

Note: Since ${a_p}$ is a term of the A.P., $p$ is a natural number. Therefore, the term $4p$ is positive. This is why we were able to add $4p$ on both sides of the inequation without changing the sign of inequality.
We used the distributive property of multiplication in the solution. The distributive property of multiplication states that $\left( {a + b} \right)\left( {c + d} \right) = a \cdot c + a \cdot d + b \cdot c + b \cdot d$.
Verification: We can verify our answer by finding the 31st and 32nd term of the A.P.
Substituting $n = 31$, $a = 121$, and $d = - 4$ in the formula for ${n^{{\text{th}}}}$ term of an A.P., we get

$$\Rightarrow {a_{31}} = 121 + \left( {31 - 1} \right)\left( { - 4} \right) \\\ \Rightarrow {a_{31}} = 121 + \left( {30} \right)\left( { - 4} \right) \\\ \Rightarrow {a_{31}} = 121 - 120 \\\ \Rightarrow {a_{31}} = 1 \\\$$

Substituting $n = 32$, $a = 121$, and $d = - 4$ in the formula for ${n^{{\text{th}}}}$ term of an A.P., we get

$$\Rightarrow {a_{32}} = 121 + \left( {32 - 1} \right)\left( { - 4} \right) \\\ \Rightarrow {a_{32}} = 121 + \left( {31} \right)\left( { - 4} \right) \\\ \Rightarrow {a_{32}} = 121 - 124 \\\ \Rightarrow {a_{32}} = - 3 \\\$$

Hence, we have verified that the 32nd term of the given A.P. is its first negative term.