Question

In an AP:
(i) Given \[a = 5,d = 3,{a_n} = 50\], find n and \[{S_n}\].
(ii) Given \[a = 7,{a_{13}} = 35\], find d and \[{S_{13}}\].
(iii) Given \[{a_{12}} = 37,d = 3\], find a and \[{S_{12}}\].
(iv) Given \[{a_3} = 15,{S_{10}} = 125\], find d and \[{a_{10}}\].
(v) Given \[d = 5,{S_9} = 75\], find a and \[{a_9}\].
(vi) Given \[a = 2,d = 8,{S_n} = 90\], find n and \[{a_n}\].
(vii) Given \[a = 8,{a_n} = 62,{S_n} = 210\], find n and d.
(viii) Given \[{a_n} = 4,d = 2,{S_n} = - 14\], find n and a.
(ix) Given \[a = 3,n = 8,S = 192\], find d.
(x) Given \[l = 28,S = 144\], and there are a total 9 terms. Find a.

Answer 1

Hint: Use the formula of Arithmetic progression sequence for the nth terms that is \[{a_n} = a + \left( {n - 1} \right)d\] where, a initial term of the AP and d is the common difference of successive numbers. Calculate the value of n. We use the formula of the sum of n terms in Arithmetic progression that is \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]. Calculate the sum of the AP \[{S_n}\].

Complete step by step answer:
Given data:
(i) \[a = 5,d = 3,{a_n} = 50\].
Now, we know about the Arithmetic progression sequence for the nth terms is:
\[{a_n} = a + \left( {n - 1} \right)d\]
Now, calculate the value of $n$. Substitute the value of \[a = 5,d = 3,{a_n} = 50\] in \[{a_n} = a + \left( {n - 1} \right)d\].
$\Rightarrow 50 = 5 + \left( {n - 1} \right)\left( 3 \right)$
$\Rightarrow 45 = 3n - 3$
$\Rightarrow 3n = 48$
$\Rightarrow n= 16$
Hence, the value of n is 16.
Now, we know about the formula of the sum of n terms in an Arithmetic progression that is:
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Simplify the above equation by substituting \[{a_n} = a + \left( {n - 1} \right)d\].
\[{S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]\]

Now, calculate the value of ${S_n}$ where \[a = 5,n = 16,{a_n} = 50\]. Substitute the values in \[{S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]\].
$\Rightarrow {S_{16}} = \dfrac{{16}}{2}\left[ {5 + 50} \right]$
$ = 8\left[ {55} \right]$
$ = 440$
Hence, the value of \[{S_n}\] is \[440\].

(ii) \[a = 7,{a_{13}} = 35\]
Now, we know about the Arithmetic progression sequence for the nth terms is:
\[{a_n} = a + \left( {n - 1} \right)d\]
Now, calculate the value of $d$. Substitute the value of \[a = 7,{a_{13}} = 35\] in \[{a_n} = a + \left( {n - 1} \right)d\].
$\Rightarrow 35 = 7 + \left( {13 - 1} \right)d$
$\Rightarrow 28 = 12d$
$\Rightarrow d = \dfrac{{28}}{{12}}$
$\Rightarrow d = \dfrac{7}{3}$
Hence, the value of d is \[\dfrac{7}{3}\].
Now, we know about the formula of the sum of n terms in Arithmetic progression that is:
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Simplify the above equation by substituting \[{a_n} = a + \left( {n - 1} \right)d\].
\[{S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]\]
Now, calculate the value of ${S_n}$ where $a = 7,{a_{13}} = 35,{\rm{ and }}d = \dfrac{7}{3}$. Substitute the values in \[{S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]\].
$\Rightarrow {S_{13}} = \dfrac{{13}}{2}\left[ {7 + 35} \right]$
$ = 6.5\left[ {42} \right]$
$ = 273$
Hence, the value of \[{S_{13}}\] is \[273\].

(iii) \[{a_{12}} = 37,d = 3\].
Now, we know about the Arithmetic progression sequence for the nth terms is:
\[{a_n} = a + \left( {n - 1} \right)d\]
Now, calculate the value of $a$. Substitute the value of \[{a_{12}} = 37,d = 3\] in \[{a_n} = a + \left( {n - 1} \right)d\].
$\Rightarrow 37 = a + \left( {12 - 1} \right)\left( 3 \right)$
$\Rightarrow 37 = a + 33$
$\Rightarrow a = 4$
Hence, the value of a is 4.
Now, we know about the formula of the sum of n terms in Arithmetic progression that is:
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Simplify the above equation by substituting \[{a_n} = a + \left( {n - 1} \right)d\].
\[{S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]\]
Now, calculate the value of ${S_n}$ where \[{a_{12}} = 37,d = 3,{\rm{ and }}a = 4\]. Substitute the values in \[{S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]\].
$\Rightarrow {S_{12}} = \dfrac{{12}}{2}\left[ {4 + 37} \right]$
$ = 6\left[ {41} \right]$
$ = 246$
Hence, the value of \[{S_{12}}\] is \[246\].

(iv) \[{a_3} = 15,{S_{10}} = 125\]
Now, we know about the Arithmetic progression sequence for the nth terms is:
\[{a_n} = a + \left( {n - 1} \right)d\]
The general formula of the AP is \[a,a + d,a + 2d,a + 3d,...\]. Now, \[{a_3} = a + 2d\] and substitute the value of \[{a_3} = 15\].
$\Rightarrow 15 = a + 2d$
$\Rightarrow a = 15 - 2d$
Now, we know about the formula of the sum of n terms in Arithmetic progression that is:
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Now, calculate the value of $a{\rm{ and }}d$. Substitute the values \[{a_3} = 15,{S_{10}} = 125,{\rm{ and }}a = 15 - 2d\] in \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\].
$\Rightarrow 125 = \dfrac{{10}}{2}\left[ {2\left( {15 - 2d} \right) + \left( {10 - 1} \right)d} \right]$
$\Rightarrow \dfrac{{125}}{5} = 30 - 4d + 9d$
$\Rightarrow 25 - 30 = 5d$
$\Rightarrow d = - 1$
Hence, the value of d is \[ - 1\].
Substitute the value \[d = - 1\] in \[a = 15 - 2d\].
$\Rightarrow a = 15 - 2\left( { - 1} \right)$
$\Rightarrow a = 17$
Hence, the value of $a$ is 17.
Now, we know about the Arithmetic progression sequence for the nth terms is:
\[{a_n} = a + \left( {n - 1} \right)d\]
Now, calculate the value of $a$. Substitute the value of \[a = 17,d = - 1,{\rm{ and }}n = 10\] in \[{a_n} = a + \left( {n - 1} \right)d\].
$\Rightarrow {a_{10}} = 17 + \left( {10 - 1} \right)\left( { - 1} \right)$
$ = 17 - 9$
$ = 8$
Hence, the value of \[{a_{10}}\] is 8.

(v) \[d = 5,{S_9} = 75\]
Now, we know about the formula of the sum of n terms in Arithmetic progression that is:
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Now, calculate the value of $a{\rm{ and }}d$. Substitute the values \[d = 5,{S_9} = 75\] in \[{S_n} = $\dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\].
$\Rightarrow 75 = \dfrac{9}{2}\left[ {2a + \left( {9 - 1} \right)5} \right]$
$\Rightarrow 75\left( {\dfrac{2}{9}} \right) = 2a + 40$
$\Rightarrow - 23.3333 = 2a$
$\Rightarrow d = - 11.6667$
Hence, the value of d is \[ - 11.6667\].
Now, we know about the Arithmetic progression sequence for the nth terms is:
\[{a_n} = a + \left( {n - 1} \right)d\]
Now, calculate the value of ${a_9}$. Substitute the value of \[d = 5,{S_9} = 75,{\rm{ and }} - 11.6667\] in \[{a_n} = a + \left( {n - 1} \right)d\].
$\Rightarrow {a_9} = - 11.6667 + \left( {9 - 1} \right)\left( 5 \right)$
$ = - 11.6667 + 40$
$= 28.3333$
Hence, the value of \[{a_9}\] is 28.3333.

(vi) \[a = 2,d = 8,{S_n} = 90\]
Now, we know about the formula of the sum of n terms in Arithmetic progression that is:
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Now, calculate the value of $a{\rm{ and }}d$. Substitute the values \[a = 2,d = 8,{\rm{ and }}{S_n} = 90\] in \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\].
$\Rightarrow 90 = \dfrac{n}{2}\left[ {2\left( 2 \right) + \left( {n - 1} \right)8} \right]$
$\Rightarrow 90 = \dfrac{n}{2}\left( {4 + 8n - 8} \right)$
$\Rightarrow 180 = - 4n + 8{n^2}$
$\Rightarrow 2{n^2} - n - 45 = 0$
On further simplification, the following is obtained:
$\Rightarrow 2{n^2} - 10n + 9n - 45 = 0$
$\Rightarrow 2n\left( {n - 5} \right) + 9\left( {n - 5} \right) = 0$
$\Rightarrow \left( {n - 5} \right)\left( {2n + 9} \right) = 0$
$\Rightarrow n = 5, - \dfrac{9}{2}$
Hence, the values of n are \[5, - \dfrac{9}{2}\].
Now, we know about the Arithmetic progression sequence for the nth terms is:
\[{a_n} = a + \left( {n - 1} \right)d\]
Now, calculate the value of ${a_n}$. Substitute the value of \[a = 2,d = 8,{\rm{ and }}n = 5\] in \[{a_n} = a + \left( {n - 1} \right)d\].
$\Rightarrow {a_5} = 2 + \left( {5 - 1} \right)\left( 8 \right)$
$ = 2 + 32$
$= 34$
Hence, the value of \[{a_5}\] is 34.

(vii) \[a = 8,{a_n} = 62,{S_n} = 210\]
Now, we know about the formula of the sum of n terms in Arithmetic progression that is:
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Simplify the above equation by substituting \[{a_n} = a + \left( {n - 1} \right)d\].
\[{S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]\]
Now, calculate the value of ${S_n}$ where \[a = 8,{a_n} = 62,{\rm{ and }}{S_n} = 210\]. Substitute the values in \[{S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]\].
$\Rightarrow 210 = \dfrac{n}{2}\left[ {8 + 62} \right]$
$\Rightarrow 420 = n\left[ {70} \right]$
$\Rightarrow n = \dfrac{{420}}{{70}}$
$\Rightarrow n= 6$
Hence, the value of \[n\] is \[6\].
Now, we know about the Arithmetic progression sequence for the nth terms is:
\[{a_n} = a + \left( {n - 1} \right)d\]
Now, calculate the value of ${a_n}$. Substitute the value of \[a = 8,{a_n} = 62,{\rm{ and }}n = 6\] in \[\Rightarrow {a_n} = a + \left( {n - 1} \right)d\].
$\Rightarrow 62 = 8 + \left( {6 - 1} \right)\left( d \right)$
$\Rightarrow 54 = 5d$
$\Rightarrow d = \dfrac{{54}}{5}$
Hence, the value of \[d\] is \[\dfrac{{54}}{5}\].

(viii) \[{a_n} = 4,d = 2,{S_n} = - 14\]
Now, we know about the formula of the sum of n terms in Arithmetic progression that is:
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Simplify the above equation by substituting \[{a_n} = a + \left( {n - 1} \right)d\].
\[{S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]\]
Now, calculate the value of ${S_n}$ where \[{a_n} = 4,d = 2,{\rm{ and }}{S_n} = - 14\]. Substitute the values in \[{S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]\].
$\Rightarrow - 14 = \dfrac{n}{2}\left[ {a + 4} \right]$
$\Rightarrow - 28 = n\left( {a + 4} \right)$
$\Rightarrow n = - \dfrac{{28}}{{a + 4}}$
Now, we know about the Arithmetic progression sequence for the nth terms is:
\[{a_n} = a + \left( {n - 1} \right)d\]
Now, calculate the value of $a$. Substitute the value of \[{a_n} = 4,d = 2,{\rm{ and }}n = - \dfrac{{28}}{{a + 4}}\] in \[{a_n} = a + \left( {n - 1} \right)d\].
$\Rightarrow 4 = a + \left( { - \dfrac{{28}}{{a + 4}} - 1} \right)\left( 2 \right)$
$\Rightarrow 4\left( {a + 4} \right) = a\left( {a + 4} \right) - 56 - 2a - 8$
$\Rightarrow 4a + 16 = {a^2} + 4a - 64 - 2a$
$\Rightarrow {a^2} - 2a - 80 = 0$
Further simplification, the following is obtained:
$\Rightarrow {a^2} - 10a + 8a - 80 = 0$
$\Rightarrow a\left( {a - 10} \right) + 8\left( {a - 10} \right) = 0$
$ \Rightarrow \left( {a - 10} \right)\left( {a + 8} \right) = 0$
$\Rightarrow a = 10, - 8$
Hence, the value of \[a\] is \[10{\rm{ and }} - 8\].
Now, substitute the value of \[a = 10{\rm{ and }} - 8\] in \[n = - \dfrac{{28}}{{a + 4}}\].
$\Rightarrow n = - \dfrac{{28}}{{10 + 4}}$
$\Rightarrow n = - 2$
$\Rightarrow n = - \dfrac{{28}}{{ - 8 + 4}}$
$\Rightarrow n= 7$
Here, the value of n cannot be negative. So, \[n = - 2\] is not possible.
Hence, the value of \[n\] is \[7\] and the value of a is \[ - 8\].

(ix) \[a = 3,n = 8,S = 192\]
Now, we know about the formula of the sum of n terms in Arithmetic progression that is:
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Now, calculate the value of $a{\rm{ and }}d$. Substitute the values \[a = 3,n = 8,{\rm{ and }}S = 192\] in \[\Rightarrow {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\].
$\Rightarrow 192 = \dfrac{8}{2}\left[ {2\left( 3 \right) + \left( {8 - 1} \right)d} \right]$
$\Rightarrow 192 = 24 + 28d$
$\Rightarrow 168 = 28d$
$\Rightarrow d = 6$
Hence, the value of d is 6.

(x) \[l = 28,S = 144\] and there are total 9 terms.
Now, we know about the formula of the sum of n terms in Arithmetic progression that is:
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Simplify the above equation by substituting \[{a_n} = a + \left( {n - 1} \right)d\]. Where, \[{a_n}\] shows the last number which can also denoted by l. that is \[{a_n} = l\]
\[\Rightarrow {S_n} = \dfrac{n}{2}\left[ {a + l} \right]\]
Now, calculate the value of ${S_n}$ where \[l = 28{\rm{ and }}S = 144\]. Substitute the values in \[{S_n} = \dfrac{n}{2}\left[ {a + l} \right]\].
$\Rightarrow 144 = \dfrac{9}{2}\left[ {a + 28} \right]$
$\Rightarrow 32 = a + 28$
$\Rightarrow a = 4$
Hence, the value of a is 4.

Note:
The general equation of the Arithmetic progression is \[a,a + d,a + 2d,a + 3d,...\], where a is the initial term of the AP and d is the common difference of successive numbers. Make sure use the formula of the sum of n terms in Arithmetic progression that is \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\] and use the Arithmetic progression sequence for the nth terms that is \[{a_n} = a + \left( {n - 1} \right)d\].