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If ${a_n} = 3 - 4n,$then show that ${a_1},{a_2},{a_3}.......$ form an A.P. Also find ${S_{20}}$.
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Hint: We will find the common difference of the series by taking any two consecutive terms let us say ${{\text{r}}^{{\text{th}}}}$ and ${(r - 1)^{th}}$ terms i.e., we’ll find the value of ${a_r} - {a_{r - 1}}$, if it comes out to be constant the given series will form an A.P. as an A.P. have a constant common difference throughout the series.
To find the sum of the first 20 terms of the series we’ll use the formula of the sum of first n terms of an A.P.
Complete step by step solution:
Given data: ${a_n} = 3 - 4n,$
Let a random term and its preceding term to find the common difference of the series
Let's take ${{\text{r}}^{{\text{th}}}}$ and ${(r - 1)^{th}}$ terms
$
{{\text{a}}_{\text{r}}}{\text{ = 3 - 4r}} \\
\Rightarrow {{\text{a}}_{{\text{r - 1}}}}{\text{ = 3 - 4(r - 1)}} \\
\Rightarrow {{\text{a}}_{{\text{r - 1}}}}{\text{ = 3 - 4r + 4}} \\
\Rightarrow {{\text{a}}_{{\text{r - 1}}}}{\text{ = 7 - 4r}} \\
$
Now common difference say d,
\[
{\text{d = }}{{\text{a}}_{\text{r}}}{\text{ - }}{{\text{a}}_{{\text{r - 1}}}} \\
\Rightarrow {\text{d = 3 - 4r - (7 - 4r)}} \\
\Rightarrow {\text{d = 3 - 4r - 7 + 4r}} \\
\Rightarrow {\text{d = - 4}} \\
\]
Putting the values of ${a_r}$ and ${a_{r - 1}}$
Since the common difference(d) is constant, it can be said that ${a_1},{a_2},{a_3}.......$ will form an A.P.
Using ${a_n} = 3 - 4n,$
Putting \[n = 1\],
$
{{\text{a}}_{\text{1}}}{\text{ = 3 - 4(1)}} \\
\therefore {{\text{a}}_{\text{1}}}{\text{ = - 1}} \\
$
Putting \[n = 2\],
$
{{\text{a}}_{\text{2}}}{\text{ = 3 - 4(2)}} \\
\therefore {{\text{a}}_{\text{2}}}{\text{ = - 5}} \\
$
Putting \[n = 3\],
$
{a_3} = 3 - 4(3) \\
\therefore {a_3} = - 9 \\
$
Therefore, the A.P. will be -1,-5,-9……….
Now, Sum of first n terms of an A.P. i.e. ${S_n}$is given by,
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Using this formula, we get,
${S_{20}} = \dfrac{{20}}{2}\left[ {2{a_1} + \left( {20 - 1} \right)( - 4)} \right]$
Using ${a_n} = 3 - 4n,$we get,
$
= 10\left[ {2(3 - 4(1)) + 19( - 4)} \right] \\
= 10\left[ {2(3 - 4) - 76} \right] \\
= 10\left[ {2( - 1) - 76} \right] \\
= 10\left[ { - 2 - 76} \right] \\
= 10\left[ { - 78} \right] \\
= - 780 \\
$
Therefore,
${S_{20}} = - 780$
Hence, the AP is -1,-5,-9,…….. and ${S_{20}} = - 780$
Note: We can also find the value of ${S_{20}}$ using the alternative formula for the sum of first n terms of an A.P. can also be written as
${S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]$
Using ${a_n} = 3 - 4n,$we get
\[
{S_{20}} = \dfrac{{20}}{2}\left[ {{a_1} + {a_{20}}} \right] \\
= \dfrac{{20}}{2}\left[ {3 - 4(1) + 3 - 4(20)} \right] \\
= 10\left[ {3 - 4 + 3 - 80} \right] \\
= 10\left[ {6 - 84} \right] \\
= 10\left[ { - 78} \right] \\
= - 780 \\
\]
To find the sum of the first 20 terms of the series we’ll use the formula of the sum of first n terms of an A.P.
Complete step by step solution:
Given data: ${a_n} = 3 - 4n,$
Let a random term and its preceding term to find the common difference of the series
Let's take ${{\text{r}}^{{\text{th}}}}$ and ${(r - 1)^{th}}$ terms
$
{{\text{a}}_{\text{r}}}{\text{ = 3 - 4r}} \\
\Rightarrow {{\text{a}}_{{\text{r - 1}}}}{\text{ = 3 - 4(r - 1)}} \\
\Rightarrow {{\text{a}}_{{\text{r - 1}}}}{\text{ = 3 - 4r + 4}} \\
\Rightarrow {{\text{a}}_{{\text{r - 1}}}}{\text{ = 7 - 4r}} \\
$
Now common difference say d,
\[
{\text{d = }}{{\text{a}}_{\text{r}}}{\text{ - }}{{\text{a}}_{{\text{r - 1}}}} \\
\Rightarrow {\text{d = 3 - 4r - (7 - 4r)}} \\
\Rightarrow {\text{d = 3 - 4r - 7 + 4r}} \\
\Rightarrow {\text{d = - 4}} \\
\]
Putting the values of ${a_r}$ and ${a_{r - 1}}$
Since the common difference(d) is constant, it can be said that ${a_1},{a_2},{a_3}.......$ will form an A.P.
Using ${a_n} = 3 - 4n,$
Putting \[n = 1\],
$
{{\text{a}}_{\text{1}}}{\text{ = 3 - 4(1)}} \\
\therefore {{\text{a}}_{\text{1}}}{\text{ = - 1}} \\
$
Putting \[n = 2\],
$
{{\text{a}}_{\text{2}}}{\text{ = 3 - 4(2)}} \\
\therefore {{\text{a}}_{\text{2}}}{\text{ = - 5}} \\
$
Putting \[n = 3\],
$
{a_3} = 3 - 4(3) \\
\therefore {a_3} = - 9 \\
$
Therefore, the A.P. will be -1,-5,-9……….
Now, Sum of first n terms of an A.P. i.e. ${S_n}$is given by,
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Using this formula, we get,
${S_{20}} = \dfrac{{20}}{2}\left[ {2{a_1} + \left( {20 - 1} \right)( - 4)} \right]$
Using ${a_n} = 3 - 4n,$we get,
$
= 10\left[ {2(3 - 4(1)) + 19( - 4)} \right] \\
= 10\left[ {2(3 - 4) - 76} \right] \\
= 10\left[ {2( - 1) - 76} \right] \\
= 10\left[ { - 2 - 76} \right] \\
= 10\left[ { - 78} \right] \\
= - 780 \\
$
Therefore,
${S_{20}} = - 780$
Hence, the AP is -1,-5,-9,…….. and ${S_{20}} = - 780$
Note: We can also find the value of ${S_{20}}$ using the alternative formula for the sum of first n terms of an A.P. can also be written as
${S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]$
Using ${a_n} = 3 - 4n,$we get
\[
{S_{20}} = \dfrac{{20}}{2}\left[ {{a_1} + {a_{20}}} \right] \\
= \dfrac{{20}}{2}\left[ {3 - 4(1) + 3 - 4(20)} \right] \\
= 10\left[ {3 - 4 + 3 - 80} \right] \\
= 10\left[ {6 - 84} \right] \\
= 10\left[ { - 78} \right] \\
= - 780 \\
\]
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