Let a be the first term and d be the common difference of the given AP.
Given, sum of 7 terms, \({S}_{7}=49\)
\(\Rightarrow\)              \(\frac {7}{2}[2a+(7-1)d]=49\)
\(\left[ \because { S }_{ n }=\frac { n }{ 2 } \left\{ 2a+\left( n-1 \right) d \right\} \right] \)
\(\Rightarrow\) \(\frac { 7 }{ 2 } \left( 2a+6d \right) =49\)
\(\Rightarrow\) a+3d= 7    ...(i)
and sum of 17 terms,\({ S }_{ 17 }=289\)
\(\Rightarrow\)   \(\frac { 17 }{ 2 } \left[ 2a+(17-1)d \right] =289\)
\(\Rightarrow\) \(\frac { 17 }{ 2 } \left[ (2a+16d) \right] =289\)
\(\Rightarrow\) a + 8d =17 ...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
 5d = 10
\(\Rightarrow\)  d = 2 
On putting d = 2 in Eq. (i), we get a + 3(2)=7​​​​​​​
\(\Rightarrow\) \(a+6=7\Rightarrow a=1\)
Now, sum of n terms \({ S }_{ n }=\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \)
\(=\frac { n }{ 2 } \left[ 2\times 1+\left( n-1 \right) 2 \right] \)
\(=\frac { n }{ 2 } \left( 2+2n-2 \right) \)
​​​​​​​\(=\frac { n }{ 2 } \left( 2+2n-2 \right) \)

a = a, d = b - a and a_n = c
a_n = a + ( n - 1 )d  \(\Rightarrow\) c = a + ( n - 1 ) ( b - a )
\(\Rightarrow\)   \({c-a\over b-a}=n-1\Rightarrow{c-a\over b-a}+1=n\)
\(\Rightarrow\)  \({c-a+b-a \over b-a}=n\Rightarrow{c+b-2a\over b-a}=n\)
\(\therefore\)   S_n = \({n\over 2}[a+{a}_{n}]\)
\(={c+b-2a\over2(b-a)}[a+c]\)
\(={(a+c)(b+v-2a)\over2(b-a)}\)

\(\because \)   \({{t}_{11} \over{t}_{18}}={2\over3}\Rightarrow{a+10d\over a+17d}={2\over3}\)
\(\Rightarrow\)  3a + 30d = 2a + 34d \(\Rightarrow\) a = 4d
and  \({{t}_{5}\over{t}_{21}}={a+4d\over a+20d}={4d+4d \over 4d+20d}\)
\(={8d\over 24d}={1\over 3}\)
\(\Rightarrow\)    t₅ : t₂₁ = 1 : 3
\(\therefore\)   \({{S}_{5}\over{S}_{21}}={{{5}\over{2}}[2a+4d]\over{{21}\over2}[2a+20d]}={5(2a+4d)\over21(2a+20d)}\)
\(={5(8d+4d)\over21[8d+20d]=}{60d\over588d}\)
\(={30\over294}={15\over147}={5\over49}\)
\(\Rightarrow\)  S₅: S₂₁ = 5:49.

(i) Money saved on Ist day = Rs 27.5
\(\because \) Sehaj increases his saving by a fixed amount of Rs 2.5
\(\therefore \) His savings forem an AP with a=27.5 and d=2.5
\(\therefore \) Money saved on 10th day
a₁₀=a + 9d = 27.5 + 9\(\times \) 2.5
= 27.5 + 22.5 = Rs 50
(ii) a₂₅=  a + 24d
= 27.5 + 24\(\times \)2.5
= 27.5 + 60 =  Rs 87.5
(iii) Total amount saved by Sehaj in 30 days
 \(=\frac { 30 }{ 2 } [2\times 27.5+(30-1)\times 2.5)]\\ =15(55+29\times 2.5)=Rs\quad 1912.5\)

Let number of sides of polygon be n.
The smallest angle = 120°.
\(\because\)  angle are in AP
\(\therefore\)  a = 120°   and common difference d = 5°
\(\therefore\)  Angles are 120°, 125°, 130°, ... up to n terms 
 Now, sum of all the interior angles
\(={n\over2}[2a+(n-1)d]\)
\(={n\over2}[2\times120+(n-1)5]\)
\(={n\over2}(235+5n)\)                              ....(i)
Also, sum of interior angles of a polygon
= ( n - 2 ) ( 180 )                             ....(ii)
From (i) and (ii), we get
\({n\over2}(235+5n)=(n-2)(180)\)
\(\Rightarrow\)   235n + 5n² = 360n - 720 
\(\Rightarrow\)  5n² - 125n + 720 = 0
\(\Rightarrow\)  5 ( n² - 25n + 144 ) = 0
\(\Rightarrow\)     n² - 25n + 144 = 0
\(\Rightarrow\)    ( n - 9 ) ( n - 16 ) = 0
\(\Rightarrow\)       n = 9 or n = 16
When n = 16,
the sixteenth angle = a + 15d
= 120 + 15 x 5 = 195°
Which cannot be an interior angle of a polygon.
\(\therefore\)  n = 9.

Let the number of days in which work was finished be n.
Number of workers on Ist day = 150
Number of workers on IInd day = 146
Number of workers on IIIrd day = 142 and so on
One day equivalent of all the worker
= 150 + 146 + 142 + ... upto to n workers
\(={n\over2}[2\times150+(n-1)\times(-4)]\)
\(={n\over2}(304-4n)=152n-{2n}^{2}\)        ....(i)
If 150 workers would have worked every day then number of days required to finish the work = ( n - 8 )
\(\therefore\)  One day equivalent of workers = 150 ( n - 8) = 150n - 1200                      ...(ii)
From (i) and (ii), we have
152n - 2n² = 150n - 1200
\(\Rightarrow\)   2n² - 2n - 1200 = 0
\(\Rightarrow\)  n² - n - 600 = .0
\(\Rightarrow\)  ( n - 25 ) ( n + 24 ) = 0
\(\Rightarrow\)  n = - 24, n = 25
\(\therefore\) Work has completed in 25 days.

T_p = a + ( p - 1 )d
\(\Rightarrow\)     \({1\over q}=a+(p-1)d\)                    ...(i)
T_q = a + ( p - 1 )d
\(\Rightarrow\)     \({1\over q}a+(q-1)d\)                         ...(ii)
Subtracting (ii) from (i), we get
\({p-q\over pq}=(p-q)d\Rightarrow d={1\over pq}\)
Putting d = \({1 \over pq}\) in (i), we have
\({1\over q}=a+{(p-q)\over pq}\Rightarrow a={1\over pq}\)
\(\therefore\)     S_pq = \({pq\over2}[2a+(pq-1)d]\)
or,      S_pq = \({pq\over2}\left[ {{2\over pq}+{(pq-1)\over pq}} \right]\)
\(={1\over 2}(pq+1)\)

Given, S₇ = 49 and S₁₇ = 289.
Let a be the first term and d be the common difference of given AP. Then,
\(\begin{aligned} S_7=49 \Rightarrow \frac{7}{2}[2 a+6 d]=49 \Rightarrow a+3 d=7 \ldots \text { (i) } \\ \end{aligned}\)
and \(\begin{aligned} S_{17}=289 \Rightarrow \frac{17}{2}[2 a+16 d]=289 \end{aligned}\)
\(\Rightarrow\) a + 8d = 17 .....(ii)
On subtracting Eq. (i) from Eq. (i), we get
5d = 10 \(\Rightarrow\) d = 2
On substituting d = 2 in Eq. (i), we get
a + 6 = 7 \(\Rightarrow\)  a = 1
Now, \(S_n=\frac{n}{2}[2 a+(n-1) d]=\frac{n}{2}[2+(n-1) 2]\)
= n[1 + n - 1] = n²

Here     a = 1, d = 1 and n = 31
S_n = \({n\over2}[2a+(n-1)d]\)
\(={31\over2}[2\times1+(31-1)]\)
\(={31\over2}[2+30]={31\over2}\times 32\)
= 496
\(\therefore\)    Piggy bank amount =  Rs 496
Amount spent = Rs 204
Amount left = Rs 100
Total pocket money = Rs 800

- 5 + ( -8 ) + (-11) + .... + (-230)
Here,    a = -5, a_n = -230
d = - 8 - ( - 5) = -3
an = a + ( n - 1) d
\(\Rightarrow\)   -230 = -5 + ( n - 1 ) ( -3 ) \(\Rightarrow\) \(\frac{-225}{-3}\) = n -1
\(\Rightarrow\)  75 + 1 = n or n = 76
\(\therefore\)  S₇₆ = \(\frac{76}{2}[-5+(-230)]=38\times(-235)=-8930\)