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# The first two terms of an A.P. are 27 and 24 respectively. How many terms of the progression are to be added to get $- 30$?A.15B.20C.25D.18

Last updated date: 12th Sep 2024
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Hint: Here, we will find the common difference and then use the formula term of $n$th term of the arithmetic progression ${T_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term and $d$ is the common difference. Apply this formula, and then substitute the value of $a$,$d$ and $n$ in the obtained equation to find the required A.P and then finally calculate the sum using the formula of sum of $n$th term of the arithmetic progression A.P., that is,${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where $a$ is the first term and $d$ is the common difference.

We are given that the first, second terms of an A.P. are 27, 24 respectively.
Since it is given that the middle form is 0 and for the middle term to exist we should have odd.
Finding the common difference $d$ by subtracting 27 from 24, we get
$\Rightarrow d = 24 - 27 \\ \Rightarrow d = - 3 \\$
Using the formula of $n$th term of the arithmetic progression A.P., that is,${T_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term and $d$ is the common difference, we get
$\Rightarrow {T_2} = a + \left( {2 - 1} \right)d \\ \Rightarrow {T_2} = a + d \\$
Substituting the value of ${T_2}$ in the left hand side of the above equation, we get
$\Rightarrow 24 = a + d$
Using the formula of sum of $n$th term of the arithmetic progression A.P., that is,${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where $a$ is the first term and $d$ is the common difference, we get
$\Rightarrow - 30 = \dfrac{n}{2}\left( {2\left( {27} \right) + \left( {n - 1} \right)\left( { - 3} \right)} \right) \\ \Rightarrow - 30 = \dfrac{n}{2}\left( {54 - 3n + 3} \right) \\ \Rightarrow - 30 = \dfrac{n}{2}\left( {57 - 3n} \right) \\$
Cross-multiplying the above equation and then rearrange the terms, we get
$\Rightarrow - 60 = n\left( {57 - 3n} \right) \\ \Rightarrow - 60 = 57n - 3{n^2} \\ \Rightarrow 3{n^2} - 57n - 60 = 0 \\$
Splitting the middle term in the left hand side of the above equation, we get
$\Rightarrow 3{n^2} - 60n + 3n - 60 = 0 \\ \Rightarrow 3n\left( {n - 20} \right) + 3\left( {n - 20} \right) = 0 \\ \Rightarrow \left( {3n + 3} \right)\left( {n - 20} \right) = 0 \\ \Rightarrow n = 20, - 1 \\$
Since the negative value is discarded, so the required value is 20.
Hence, option B is correct.

Note: In solving these types of questions, you should be familiar with the formula of sum of the arithmetic progression and their sums. Some students use the formula of sum, $S = \dfrac{n}{2}\left( {a + l} \right)$, where $l$ is the last term, but have the to find the value of ${a_n}$ , so it will be wrong. We can also find the value of $n$th term by find the value of ${S_n} - {S_{n - 1}}$, where ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where $a$ is the first term and $d$ is the common difference. But this is a longer method, which takes time, so we will use the above method. One should know the ${a_n}$ is the $n$th term in the arithmetic progression.