# If $x < - 1$ and $x,{\text{ }}\left| {x + 1} \right|,{\text{ }}\left| {x - 1} \right|$are in A.P. Then the sum of the first twenty terms of the A.P.

$

a.{\text{ 150}} \\

b.{\text{ 280}} \\

c.{\text{ 350}} \\

d.{\text{ 180}} \\

$

Last updated date: 26th Mar 2023

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Answer

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Hint – If a number is less than zero than the modulus of that number is negative, the sum of n terms of an A.P is given as ${S_n} = \dfrac{n}{2}\left( {2{a_1} + \left( {n - 1} \right)d} \right)$ use this property to reach the answer.

It is given that $x < - 1$

$ \Rightarrow x + 1 < 0$

As we know If a number is less than zero than the modulus of that number is negative.

$ \Rightarrow \left| {x + 1} \right| = - \left( {x + 1} \right) = - x - 1$………… (2)

Again $x < - 1$

Subtract 1 on both sides in the above equation.

$

\Rightarrow x - 1 < - 1 - 1 \\

\Rightarrow x - 1 < - 2 \\

$

So, (x - 1) is also less than zero.

$ \Rightarrow \left| {x - 1} \right| = - \left( {x - 1} \right) = - x + 1$………….. (3)

Now it is given that $x,{\text{ }}\left| {x + 1} \right|,{\text{ }}\left| {x - 1} \right|$these are in A.P.

So from equation (2) and (3)

$ \Rightarrow x,{\text{ }}\left( { - x - 1} \right),{\text{ }}\left( { - x + 1} \right)$ Forms an A.P.

So according to A.P property which is twice of second term is equal to sum of first and third term

$

\Rightarrow 2\left( { - x - 1} \right) = x + \left( { - x + 1} \right) \\

\Rightarrow - 2x - 2 = x - x + 1 \\

\Rightarrow - 2x = 3 \\

\Rightarrow x = \dfrac{{ - 3}}{2} \\

$

So the series becomes

\[

x,{\text{ }}\left| {x + 1} \right|,{\text{ }}\left| {x - 1} \right| \\

\dfrac{{ - 3}}{2},\left| {\dfrac{{ - 3}}{2} + 1} \right|,\left| {\dfrac{{ - 3}}{2} - 1} \right|............... \\

\dfrac{{ - 3}}{2},\left| {\dfrac{{ - 1}}{2}} \right|,\left| {\dfrac{{ - 5}}{2}} \right|............... \\

\dfrac{{ - 3}}{2},\dfrac{1}{2},\dfrac{5}{2},................ \\

\]

Because modulus of any negative number is positive.

So the common difference of this A.P is

\[d = \dfrac{1}{2} - \dfrac{{ - 3}}{2} = \dfrac{1}{2} + \dfrac{3}{2} = \dfrac{4}{2} = 2\]

Now we have to find out the sum of 20 terms, so using the formula of sum of an A.P

${S_n} = \dfrac{n}{2}\left( {2{a_1} + \left( {n - 1} \right)d} \right)$, where n is number of terms, d is common difference and ${a_1}$ is the first term.

So, n = 20, d = 2, ${a_1} = \dfrac{{ - 3}}{2}$

$

\Rightarrow {S_{20}} = \dfrac{{20}}{2}\left( {2\left( {\dfrac{{ - 3}}{2}} \right) + \left( {20 - 1} \right)2} \right) \\

\Rightarrow {S_{20}} = 10\left( { - 3 + 38} \right) = 10\left( {35} \right) = 350 \\

$

So this is the required sum of the A.P

Hence, option (c) is correct.

Note – in such types of questions always remember the property of modulus and the basic properties of an A.P which is stated above then using these properties first determine the series, then determine the common difference then using the formula of sum of an A.P calculate the sum of 20 terms which is the required answer.

It is given that $x < - 1$

$ \Rightarrow x + 1 < 0$

As we know If a number is less than zero than the modulus of that number is negative.

$ \Rightarrow \left| {x + 1} \right| = - \left( {x + 1} \right) = - x - 1$………… (2)

Again $x < - 1$

Subtract 1 on both sides in the above equation.

$

\Rightarrow x - 1 < - 1 - 1 \\

\Rightarrow x - 1 < - 2 \\

$

So, (x - 1) is also less than zero.

$ \Rightarrow \left| {x - 1} \right| = - \left( {x - 1} \right) = - x + 1$………….. (3)

Now it is given that $x,{\text{ }}\left| {x + 1} \right|,{\text{ }}\left| {x - 1} \right|$these are in A.P.

So from equation (2) and (3)

$ \Rightarrow x,{\text{ }}\left( { - x - 1} \right),{\text{ }}\left( { - x + 1} \right)$ Forms an A.P.

So according to A.P property which is twice of second term is equal to sum of first and third term

$

\Rightarrow 2\left( { - x - 1} \right) = x + \left( { - x + 1} \right) \\

\Rightarrow - 2x - 2 = x - x + 1 \\

\Rightarrow - 2x = 3 \\

\Rightarrow x = \dfrac{{ - 3}}{2} \\

$

So the series becomes

\[

x,{\text{ }}\left| {x + 1} \right|,{\text{ }}\left| {x - 1} \right| \\

\dfrac{{ - 3}}{2},\left| {\dfrac{{ - 3}}{2} + 1} \right|,\left| {\dfrac{{ - 3}}{2} - 1} \right|............... \\

\dfrac{{ - 3}}{2},\left| {\dfrac{{ - 1}}{2}} \right|,\left| {\dfrac{{ - 5}}{2}} \right|............... \\

\dfrac{{ - 3}}{2},\dfrac{1}{2},\dfrac{5}{2},................ \\

\]

Because modulus of any negative number is positive.

So the common difference of this A.P is

\[d = \dfrac{1}{2} - \dfrac{{ - 3}}{2} = \dfrac{1}{2} + \dfrac{3}{2} = \dfrac{4}{2} = 2\]

Now we have to find out the sum of 20 terms, so using the formula of sum of an A.P

${S_n} = \dfrac{n}{2}\left( {2{a_1} + \left( {n - 1} \right)d} \right)$, where n is number of terms, d is common difference and ${a_1}$ is the first term.

So, n = 20, d = 2, ${a_1} = \dfrac{{ - 3}}{2}$

$

\Rightarrow {S_{20}} = \dfrac{{20}}{2}\left( {2\left( {\dfrac{{ - 3}}{2}} \right) + \left( {20 - 1} \right)2} \right) \\

\Rightarrow {S_{20}} = 10\left( { - 3 + 38} \right) = 10\left( {35} \right) = 350 \\

$

So this is the required sum of the A.P

Hence, option (c) is correct.

Note – in such types of questions always remember the property of modulus and the basic properties of an A.P which is stated above then using these properties first determine the series, then determine the common difference then using the formula of sum of an A.P calculate the sum of 20 terms which is the required answer.

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