Chapter 5: Arithmetic Progressions

Ans: Given, the first Term, $a=2$  ….. (1)

Given, the common Difference, \[d=7-2=5\] …..(2)

Given, the number of Terms, \[n=10\]  …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, ${{S}_{n}}=\dfrac{10}{2}\left[ 2\left( 2 \right)+\left( 10-1 \right)\left( 5 \right) \right]$

$\Rightarrow {{S}_{n}}=5\left[ 4+45 \right]$

$\therefore {{S}_{n}}=245$ 

Ans: Given, the first Term, $a=-37$  ….. (1)

Given, the common Difference, \[d=-33-\left( -37 \right)=4\] …..(2)

Given, the number of Terms, \[n=12\]  …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, ${{S}_{n}}=\dfrac{12}{2}\left[ 2\left( -37 \right)+\left( 12-1 \right)\left( 4 \right) \right]$

$\Rightarrow {{S}_{n}}=6\left[ -74+44 \right]$

$\therefore {{S}_{n}}=-180$ 

Ans: Given, the first Term, $a=0.6$  ….. (1)

Given, the common Difference, \[d=1.7-0.6=1.1\] …..(2)

Given, the number of Terms, \[n=100\]  …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, ${{S}_{n}}=\dfrac{100}{2}\left[ 2\left( 0.6 \right)+\left( 100-1 \right)\left( 1.1 \right) \right]$

$\Rightarrow {{S}_{n}}=50\left[ 1.2+108.9 \right]$

$\therefore {{S}_{n}}=5505$ 

Ans: Given, the first Term, $a=\dfrac{1}{15}$  ….. (1)

Given, the common Difference, \[d=\dfrac{1}{12}-\dfrac{1}{15}=\dfrac{1}{60}\] …..(2)

Given, the number of Terms, \[n=11\]  …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, ${{S}_{n}}=\dfrac{11}{2}\left[ 2\left( \dfrac{1}{15} \right)+\left( 11-1 \right)\left( \dfrac{1}{60} \right) \right]$

$\Rightarrow {{S}_{n}}=\dfrac{11}{2}\left[ \dfrac{4+5}{30} \right]$

\[\therefore {{S}_{n}}=\dfrac{33}{20}\] 

Ans: Given, the first Term, $a=7$  ….. (1)

Given, the common Difference, \[d=10\dfrac{1}{2}-7=\dfrac{7}{2}\] …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$   ….. (3)

Substituting the values from (1) and (2) in (3) we get, 

${{a}_{n}}=7+\dfrac{7}{2}\left( n-1 \right)=\dfrac{7}{2}\left( n+1 \right)$  ….. (4)

Given, last term of the series, \[{{a}_{n}}=84\]  …..(5)

Substituting (5) in (4) we get, $84=\dfrac{7}{2}\left( n+1 \right)$

$\Rightarrow 24=\left( n+1 \right)$ 

\[\therefore n=23\]  ……(6)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$     …..(7)

Substituting the values from (1), (5) and (6) in (7) we get, ${{S}_{n}}=\dfrac{23}{2}\left[ 7+84 \right]$

$\Rightarrow {{S}_{n}}=\dfrac{23}{2}\left( 91 \right)$

\[\therefore {{S}_{n}}=1046\dfrac{1}{2}\] 

Ans: Given, the first Term, $a=34$  ….. (1)

Given, the common Difference, \[d=32-34=-2\] …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$   ….. (3)

Substituting the values from (1) and (2) in (3) we get, 

${{a}_{n}}=34-2\left( n-1 \right)=36-2n$  ….. (4)

Given, last term of the series, \[{{a}_{n}}=10\]  …..(5)

Substituting (5) in (4) we get, $10=36-2n$

$\Rightarrow 2n=26$ 

\[\therefore n=13\]  ……(6)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$     …..(7)

Substituting the values from (1), (5) and (6) in (7) we get, ${{S}_{n}}=\dfrac{13}{2}\left[ 34+10 \right]$

$\Rightarrow {{S}_{n}}=\dfrac{13}{2}\left( 44 \right)$

\[\therefore {{S}_{n}}=286\] 

Ans: Given, the first Term, $a=-5$  ….. (1)

Given, the common Difference, \[d=-8-\left( -5 \right)=-3\] …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$   ….. (3)

Substituting the values from (1) and (2) in (3) we get, 

${{a}_{n}}=-5-3\left( n-1 \right)=-2-3n$  ….. (4)

Given, last term of the series, \[{{a}_{n}}=-230\]  …..(5)

Substituting (5) in (4) we get, $-230=-2-3n$

$\Rightarrow -228=-3n$ 

\[\therefore n=76\]  ……(6)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$     …..(7)

Substituting the values from (1), (5) and (6) in (7) we get, ${{S}_{n}}=\dfrac{76}{2}\left[ -5+\left( -230 \right) \right]$

$\Rightarrow {{S}_{n}}=\dfrac{76}{2}\left( -235 \right)$

\[\therefore {{S}_{n}}=-8930\] 

Ans: Given, the first Term, $a=5$  ….. (1)

Given, the common Difference, \[d=3\] …..(2)

Given, ${{n}^{th}}$ term of the A.P., \[{{\mathbf{a}}_{\mathbf{n}}}=\mathbf{50}\]  …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$   ….. (4)

Substituting the values from (1), (2) and (3) in (4) we get, 

$50=5+3\left( n-1 \right)=2+3n$ 

Simplifying it further we get, 

$n=\dfrac{50-2}{3}$ 

$\therefore n=16$   …..(5)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     …..(6)

Substituting the values from (1), (2) and (5) in (6) we get, ${{S}_{n}}=\dfrac{16}{2}\left[ 2\left( 5 \right)+\left( 16-1 \right)\left( 3 \right) \right]$

$\Rightarrow {{S}_{n}}=8\left[ 10+45 \right]$

\[\therefore {{S}_{n}}=440\] 

Ans: Given, the first Term, $a=7$  ….. (1)

Given, ${{13}^{th}}$ term of the A.P., \[{{a}_{13}}=35\]  …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$   ….. (3)

Substituting the values from (1), (2) in (3) we get, 

$35=7+\left( 13-1 \right)d=7+12d$ 

Simplifying it further we get, 

$d=\dfrac{28}{12}$ 

\[\therefore d=\dfrac{7}{3}\]   …..(4)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     …..(5)

Substituting the values from (1) and (4) in (5) we get, ${{S}_{13}}=\dfrac{13}{2}\left[ 2\left( 7 \right)+\left( 13-1 \right)\left( \dfrac{7}{3} \right) \right]$

$\Rightarrow {{S}_{13}}=\dfrac{13}{2}\left[ 14+28 \right]$

\[\therefore {{S}_{13}}=273\] 

Ans: Given, the common difference, $d=3$  ….. (1)

Given, ${{12}^{th}}$ term of the A.P., \[{{\mathbf{a}}_{12}}=37\]  …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$   ….. (3)

Substituting the values from (1), (2) in (3) we get, 

$37=a+3\left( 12-1 \right)=a+33$ 

Simplifying it further we get,  

\[\therefore a=4\]   …..(4)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     …..(5)

Substituting the values from (1) and (4) in (5) we get, ${{S}_{12}}=\dfrac{12}{2}\left[ 2\left( 4 \right)+\left( 12-1 \right)\left( 3 \right) \right]$

$\Rightarrow {{S}_{12}}=6\left[ 8+33 \right]$

\[\therefore {{S}_{12}}=246\]

Ans: Given, ${{3}^{rd}}$ term of the A.P., \[{{\mathbf{a}}_{3}}=1\mathbf{5}\]  …..(1)

Given, the sum of terms, \[{{\mathbf{S}}_{10}}=125\]  ….. (2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$   ….. (3)

Substituting the values from (1) in (3) we get, 

$15=a+\left( 3-1 \right)d=a+2d$  …..(4)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     …..(5)

Substituting the values from (1) in (5) we get, $125=\dfrac{10}{2}\left[ 2a+\left( 10-1 \right)d \right]$

$\Rightarrow 125=5\left[ 2a+9d \right]$

\[\therefore 25=2a+9d\]  …..(5)

Let us solve equations (4) and (5) by subtracting twice of (4) from (5) we get,

\[25-30=\left( 2a+9d \right)-\left( 2a+4d \right)\]

$\Rightarrow -5=5d$ 

$\therefore d=-1$    …..(6)

From (4) and (6) we get, $a=17$   …..(7)

From (3), (6) and (7) for $n=10$ we get,

${{a}_{10}}=17-\left( 10-1 \right)$

$\therefore {{a}_{10}}=8$  

Ans: Given, common difference, \[d=5\]  …..(1)

Given, the sum of terms, \[{{\mathbf{S}}_{9}}=75\]  ….. (2)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     …..(3)

Substituting the values from (1), (2) in (3) we get, $75=\dfrac{9}{2}\left[ 2a+5\left( 9-1 \right) \right]$

$\Rightarrow 25=3\left[ a+20 \right]$

$\Rightarrow 3a=-35$

\[\therefore a=-\dfrac{35}{3}\]  …..(4)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$   ….. (5)

Substituting the values from (1), (4) in (5) we get, 

${{a}_{9}}=-\dfrac{35}{3}+5\left( 9-1 \right)$ 

$\Rightarrow {{a}_{9}}=-\dfrac{35}{3}+40$ 

$\therefore {{a}_{9}}=\dfrac{85}{3}$ 

Ans: Given, common difference, \[d=8\]  …..(1)

Given, first term, \[a=2\]  …..(2)

Given, the sum of terms, \[{{S}_{\mathbf{n}}}=9\mathbf{0}\]  ….. (3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     …..(3)

Substituting the values from (1), (2), (3) in (4) we get, $90=\dfrac{n}{2}\left[ 2\left( 2 \right)+8\left( n-1 \right) \right]$

$\Rightarrow 45=n\left[ 2n-1 \right]$

$\Rightarrow 2{{n}^{2}}-n-45=0$

$\Rightarrow 2{{n}^{2}}-10n+9n-45=0$

$\Rightarrow 2n\left( n-5 \right)+9\left( n-5 \right)=0$

$\Rightarrow \left( n-5 \right)\left( 2n+9 \right)=0$

\[\therefore n=5\]  …..(4)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$   ….. (5)

Substituting the values from (1), (2), (4) in (5) we get, 

${{a}_{5}}=2+8\left( 5-1 \right)$ 

$\Rightarrow {{a}_{5}}=2+32$ 

$\therefore {{a}_{5}}=34$ 

Ans: Given, first term, \[a=8\]  …..(1)

Given, the sum of terms, \[{{S}_{\mathbf{n}}}=21\mathbf{0}\]  ….. (2)

Given, the ${{n}^{th}}$ term, \[{{\mathbf{a}}_{\mathbf{n}}}=62\] …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     …..(4)

Substituting the values from (1), (2) in (4) we get, $210=\dfrac{n}{2}\left[ 2\left( 8 \right)+d\left( n-1 \right) \right]$

$\Rightarrow 420=n\left[ 16+\left( n-1 \right)d \right]$  …..(4)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$   ….. (5)

Substituting the values from (1), (3) in (5) we get, 

$62=8+\left( n-1 \right)d$   …..(6) 

Let us solve equations (4) and (6) by subtracting $n$ times of (6) from (4) we get,

$420-62n=\left( 16n+n\left( n-1 \right)d \right)-\left( 8n+n\left( n-1 \right)d \right)$

$\Rightarrow 420-62n=8n$ 

$\Rightarrow 420=70n$

$\therefore n=6$  ……(7)

Substituting the values from (7) in (6) we get, 

 $62=8+\left( 6-1 \right)d$

$\Rightarrow 54=5d$ 

$\therefore d=\dfrac{54}{5}$ 

Ans: Given, common difference, \[d=2\]  …..(1)

Given, the sum of terms, \[{{S}_{\mathbf{n}}}=-14\]  ….. (2)

Given, the ${{n}^{th}}$ term, \[{{\mathbf{a}}_{\mathbf{n}}}=4\] …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     …..(4)

Substituting the values from (1), (2) in (4) we get, $-14=\dfrac{n}{2}\left[ 2a+2\left( n-1 \right) \right]$

$\Rightarrow -14=n\left[ a+n-1 \right]$  …..(5)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$   ….. (6)

Substituting the values from (1), (3) in (6) we get, 

$4=a+2\left( n-1 \right)$   …..(7) 

Let us solve equations (5) and (7) by substituting the value of $a$ from (7) in (5) we get,

$-14=n\left[ \left( 4-2\left( n-1 \right) \right)+n-1 \right]$

\[\Rightarrow -14=n\left[ 5-n \right]\]

$\Rightarrow {{n}^{2}}-5n-14=0$ 

$\Rightarrow {{n}^{2}}-7n+2n-14=0$

$\Rightarrow \left( n-7 \right)\left( n+2 \right)=0$

$\therefore n=7$ (Since $n$ cannot be negative)  ……(8)

Substituting the values from (8) in (7) we get, 

 $4=a+2\left( 7-1 \right)$

$\Rightarrow 4=a+12$ 

$\therefore a=-8$ 

Ans: Given, first term, \[a=3\]  …..(1)

Given, the sum of terms, \[{{S}_{\mathbf{n}}}=192\]  ….. (2)

Given, the number of terms, \[n=8\] …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     …..(4)

Substituting the values from (1), (2) in (4) we get, $192=\dfrac{8}{2}\left[ 2\left( 3 \right)+d\left( 8-1 \right) \right]$

$\Rightarrow 192=4\left[ 6+7d \right]$

$\Rightarrow 48=6+7d$

$\Rightarrow 42=7d$

$\therefore d=6$ 

Ans: Given, last term, \[l=28\]  …..(1)

Given, the sum of terms, \[{{S}_{\mathbf{n}}}=144\]  ….. (2)

Given, the number of terms, \[n=9\] …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$     …..(4)

Substituting the values from (1), (2) in (4) we get, $144=\dfrac{9}{2}\left[ a+28 \right]$

$\Rightarrow 32=a+28$

$\therefore a=4$ 

Ans: Given, common difference, \[d=17-9=8\]  …..(1)

Given, first term, \[a=9\]  …..(2)

Given, the sum of terms, \[{{S}_{\mathbf{n}}}=636\]  ….. (3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     …..(3)

Substituting the values from (1), (2), (3) in (4) we get, $636=\dfrac{n}{2}\left[ 2\left( 9 \right)+8\left( n-1 \right) \right]$

$\Rightarrow 636=n\left( 5+4n \right)$

$\Rightarrow 4{{n}^{2}}+5n-636=0$

$\Rightarrow 4{{n}^{2}}+53n-48n-636=0$

$\Rightarrow n\left( 4n+53 \right)-12\left( 4n+53 \right)=0$

$\Rightarrow \left( n-12 \right)\left( 4n+53 \right)=0$

$\Rightarrow n=12\text{ }or\text{ }-\dfrac{53}{4}$

Since $n$ can only be a natural number \[\therefore n=12\]

Ans: Given, first term, \[a=5\]  …..(1)

Given, the sum of terms, \[{{S}_{\mathbf{n}}}=400\]  ….. (2)

Given, the ${{n}^{th}}$ term, \[{{\mathbf{a}}_{\mathbf{n}}}=45\] …..(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$     …..(4)

Substituting the values from (1), (2), (3) in (4) we get, $400=\dfrac{n}{2}\left[ 5+45 \right]$

$\Rightarrow 400=25n$ 

$\therefore n=16$ 

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$   ….. (5)

Substituting the values from (1), (3) in (5) we get, 

$45=5+\left( 16-1 \right)d$   

$\Rightarrow 40=15d$ 

$\therefore d=\dfrac{8}{3}$

Ans: Given, first term, \[a=17\]  …..(1)

Given, the common difference, \[d=9\]  ….. (2)

Given, the ${{n}^{th}}$ term, \[{{\mathbf{a}}_{\mathbf{n}}}=350\] …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$   ….. (4)

Substituting the values from (1), (2), (3) in (4) we get, 

$350=17+9\left( n-1 \right)$   

$\Rightarrow 333=9\left( n-1 \right)$ 

$\Rightarrow 37=\left( n-1 \right)$

$\therefore n=38$ ……(5)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$     …..(6)

Substituting the values from (1), (5), (3) in (6) we get, ${{S}_{38}}=\dfrac{38}{2}\left[ 17+350 \right]$

$\Rightarrow {{S}_{38}}=19\left( 367 \right)$ 

$\therefore {{S}_{38}}=6973$ 

Ans: Given, the common difference, \[d=7\]  ….. (1)

Given, the ${{22}^{nd}}$ term, \[{{\mathbf{a}}_{22}}=149\] …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$   ….. (3)

Substituting the values from (1), (2) in (3) we get, 

$149=a+7\left( 22-1 \right)$   

$\Rightarrow 149=a+147$ 

$\therefore a=2$  ……(4)

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$     …..(5)

Substituting the values from (1), (2), (4) in (5) we get, ${{S}_{22}}=\dfrac{22}{2}\left[ 2+149 \right]$

$\Rightarrow {{S}_{22}}=11\left( 151 \right)$ 

$\therefore {{S}_{22}}=1661$ 

Ans: Given, the ${{2}^{nd}}$ term, \[{{\mathbf{a}}_{2}}=14\]  ….. (1)

Given, the ${{3}^{rd}}$ term, \[{{\mathbf{a}}_{3}}=18\] …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$   ….. (3)

Substituting the values from (1) in (3) we get, 

$14=a+d$   …..(4)

Substituting the values from (2) in (3) we get, 

$18=a+2d$   …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

$18-14=\left( a+2d \right)-\left( a+d \right)$ 

$\therefore d=4$  ……(6)

Substituting the value from (6) in (4) we get $a=10$.   …..(7) 

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     …..(8)

Substituting the values from (7), (6) in (8) we get for $n=51$ ,

${{S}_{51}}=\dfrac{51}{2}\left[ 2\left( 10 \right)+4\left( 51-1 \right) \right]$

$\Rightarrow {{S}_{51}}=\dfrac{51}{2}\left[ 20+200 \right]$ 

$\therefore {{S}_{51}}=5610$ 

Ans: Given, the sum of first $7$ terms, \[{{S}_{7}}=49\]  ….. (1)

Given, the sum of first $17$ terms, \[{{S}_{17}}=289\]   …..(2)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     ….. (3)

Substituting the values from (1) in (3) we get, 

\[49=\dfrac{7}{2}\left[ 2a+\left( 7-1 \right)d \right]\]

\[\Rightarrow 7=a+3d\]   …..(4)

Substituting the values from (2) in (3) we get, 

\[289=\dfrac{17}{2}\left[ 2a+\left( 17-1 \right)d \right]\]

\[\Rightarrow 17=a+8d\]   …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

$17-7=\left( a+8d \right)-\left( a+3d \right)$ 

$\Rightarrow 10=5d$ 

$\therefore d=2$  ……(6)

Substituting the value from (6) in (4) we get $a=1$.   …..(7) 

Substituting the values from (7), (6) in (3) we get,

 ${{S}_{n}}=\dfrac{n}{2}\left[ 2+2\left( n-1 \right) \right]$

$\therefore {{S}_{n}}={{n}^{2}}$ 

Ans:  Consider two consecutive terms of the given sequence. Say ${{a}_{n}},{{a}_{n+1}}$. Difference between these terms will be 

${{a}_{n+1}}-{{a}_{n}}=\left[ 3+4\left( n+1 \right) \right]-\left[ 3+4n \right]$ 

$\Rightarrow {{a}_{n+1}}-{{a}_{n}}=4\left( n+1 \right)-4n$ 

$\Rightarrow {{a}_{n+1}}-{{a}_{n}}=4$

Which is a constant $\forall n\in \mathbb{N}$. 

For $n=1$, ${{a}_{1}}=3+4=7$ 

Therefore, it is an A.P. with first term $7$ and common difference $4$. 

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     

Therefore, ${{S}_{15}}=\dfrac{15}{2}\left[ 2\left( 7 \right)+4\left( 15-1 \right) \right]$

$\Rightarrow {{S}_{15}}=\dfrac{15}{2}\left[ 14\left( 5 \right) \right]$ 

$\therefore {{S}_{15}}=525$ 

Ans: Consider two consecutive terms of the given sequence. Say ${{a}_{n}},{{a}_{n+1}}$. Difference between these terms will be 

${{a}_{n+1}}-{{a}_{n}}=\left[ 9-5\left( n+1 \right) \right]-\left[ 9-5n \right]$ 

$\Rightarrow {{a}_{n+1}}-{{a}_{n}}=-5\left( n+1 \right)+5n$ 

$\Rightarrow {{a}_{n+1}}-{{a}_{n}}=-5$

Which is a constant $\forall n\in \mathbb{N}$. 

For $n=1$, ${{a}_{1}}=9-5=4$ 

Therefore, it is an A.P. with first term $4$ and common difference $-5$. 

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     

Therefore, ${{S}_{15}}=\dfrac{15}{2}\left[ 2\left( 4 \right)-5\left( 15-1 \right) \right]$

\[\Rightarrow {{S}_{15}}=15\left[ -31 \right]\] 

$\therefore {{S}_{15}}=-465$