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

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 ${a_n} = a + \left( {n - 1} \right)d$.