1.   Find the sum of the following APs:
(i)  2, 7, 12, . . ., to 10 terms. 
Solution: 
	Since sum of the first n terms of an AP is given by
	$S_n$=$n/2$$[2a+(n-1)d]$
	Given: $a=2$;$a_2=7$;
	So, $d=7-2=5$
	For sum of first 10 terms is 
	$S_10$=$10/2$$[2(2)+(10-1)5]$
	$S_10$=$5$$[4+45]$
	$S_10$=$245$
	
 Ans:Sum of given AP is 245  

(ii)  –37, –33, –29, . . ., to 12 terms. 
Solution:
	Given: $a=-37$;$a_2=-33$,
	So, $d=a_2-a=4$
	For sum of first 12 terms is 
	$S_12$=$12/2$$[2(-37)+(12-1)4]$
	$S_12$=$6$$[-74+44]$
	$S_12$=$180$
	
	Ans:Sum of given AP is 180 

(iii)  0.6, 1.7, 2.8, . . ., to 100 terms.
    Solution:
	Given: $a=0.6$;$a_2=1.7$;$a_3=2.8$
	Since $a_2-a$=$a_3-a_2$=$1.7-0.6=2.8-1.7$
	So, the given series is in AP with common difference $d=1.1$
	Since, sum of first 100 terms is given as
	$S_n$=$n/2$$[2a+(n-1)d]$
	$S_100$=$100/2$$[2(0.6)+(100-1)1.1]$
	$S_100$=$50$$[1.2+108.9]$
	$S_100$=$5505$
	
	Ans:Sum of given AP is 5505 

          
(iv) $1/15$ , $1/12$,$1/10$, . . ., to 11 terms. 
     Solution:
	 Given: $a=1/15$;$a_2=1/12$;$a_3=1/10$
	 Since $a_2-a=1/12-1/15=1/60$
	 and   $a_3-a_2=1/10-1/12=1/60$
	 Since the common difference is same , so the numbers are in AP.
	 Since, sum of 11 terms is given as
	 $S_11$=$11/2$$[2(1/15)+(11-1)(1/60)]$
	 $S_11$=$11/2$$[2/15+1/6]$
	 $S_11$=$11/2$$[2/15+1/6]$
	 $S_11$=$99/60$
	 
2.   Find the sums given below :
(i) 7+10$1/2$+14 + . . . + 84
Solution:
Since the first term is $7$ and the last term is $84$, 
we have to find whether the given numbers are in AP and 
then find the number of terms in the given AP to find the sums of given numbers.

Given: $a=7$;$a_2=21/2$;$a_3=14$;...;$a_n=84$
For a numbers to be in AP's ;$a_{k+1}-a_k=d$
Therefore $a_2-a=21/2-7=7/2$ 
          $a_3-a_2=14-21/2=7/2$
	Since the common difference between the consecutive numbers are same,
	the numbers are in AP.
     Given: $a_n=84$
	 Since $a_n=a+(n-1)d$
	 $84=7+(n-1)7/2$
	 $n-1=(77)2/7=22$
	 $n=23$
	 There are total 23 terms in the series.
Method 1:	 So, sum of numbers in AP upto nth terms is given as
	 $S_n$=$n/2$$[2a+(n-1)d]$
	 $S_23$=$23/2$$[2(7)+(23-1)(7/2)]$
	 $S_23=1046$$1/2$
	 
Method 2:
    $S_n=n/2(a+l)$
	where $a$ is the first term and $l$ is the last term.
	$S_23=23/2(7+84)$
	$S_23=2093/2$
	$S_23=1046$$1/2$

Ans: Sum of given numbers is $1046$$1/2$ 

(ii)34 + 32 + 30 + . . . + 10
Solution:
Since the first term is $34$ and the last term is $10$, 
we have to find whether the given numbers are in AP and 
then find the number of terms in the given AP to find the sums of given numbers.
Given: $a=34$;$a_2=32$;$a_3=30$;...;$a_n=10$
For a numbers to be in AP's ;$a_{k+1}-a_k=d$
 $a_2-a=32-34=-2$
 $a_3-a_2=30-32=-2$
Since the common difference between the consecutive numbers are same,
the numbers are in AP's with $d=-2$

Now find the numbers of terms in the series.
Given $a_n=10$
$a_n=a+(n-1)d$
$10=34+(n-1)(-2)$
$(n-1)=24/2$
$n=13$
So,total number of term in the series is 13.

Since sum of $n$ number in the AP is given by
	$S_n=n/2(a+l)$
	where $a$ is the first term and $l$ is the last term.
	$S=13/2(34+10)$
	$S=286$
Ans: Sum of given numbers is 286 	

(iii)–5 + (–8) + (–11) + . . . + (–230)

Solution:
Since the first term is $-5$ and the last term is $-230$, 
we have to find whether the given numbers are in AP and 
then find the number of terms in the given AP to find the sums of given numbers.
Given: $a=-5$;$a_2=-8$;$a_3=-11$;...;$a_n=-230$
For a numbers to be in AP's ;$a_{k+1}-a_k=d$
 $a_2-a=-8-(-5)=-3$
 $a_3-a_2=-11-(-8)=-3$
Since the common difference between the consecutive numbers are same,
the numbers are in AP's with $d=-3$

Now find the numbers of terms in the series.
Given $a_n=-230$
$-230=-5+(n-1)(-3)$
$3(n-1)=225$
$n-1=75$
$n=76$
So,total number of term in the series is 76.

Since sum of $n$ number in the AP is given by
	$S_n=n/2(a+l)$
	where $a$ is the first term and $l$ is the last term.
	$S=76/2(-5+(-230))$
	$S=-8930$
Ans: Sum of given numbers is -8930 

3.   In an AP:
(i)  given $a$ = 5, $d$ = 3, $a_n$ = 50, find $n$ and Sn.
Solution:
Since $a_n=a+(n-1)d$
      $50=5+(n-1)3$
	  $n-1=15$
	  $n=16$
Since 
    $S_n=n/2(a+l)$
	    $=16/2(5+50)$
		$=440$
Ans: $n=16$ and $S_n=440$
	  
	  
(ii)  given $a$ = 7,$a_{13}$= 35,  find $d$ and $S_{13}$ . 
Solution:
	Since $a_n=a+(n-1)d$
	      $a_13=a+(13-1)d$
	      $35=7+(12)d$
	      $28/12=d$
		  $d=7/3$
		  
	Method 1:	  
	$S_n=n/2(a+l)$
	$S_13=13/2(7+35)$
	$S_13=13/2(42)$
	$S_13=273$
	
	Method 2:
	$S_n=n/2[{2a+(n-1)d}]$
	$S_13=13/2[{2(7)+(13-1)(7/3)}]$
	$S_13=13/2[{14+28}]$
	$S_13=13/2(42)$
	$S_13=273$
Ans: $d=7/3$ and $S_13=273$

(iii)  given $a_{12}$ = 37, $d$ = 3, find $a$ and $S_{12}$  . 
Solution:
	 $a_12=a+(12-1)3$
	 $37=a+11(3)$
	 $a=4$
	 
	 $S_n=n/2(a+l)$
	 $S_12=12/2(4+37)$
	 $S_12=6(41)$
	 $S_12=246$
Ans: $a=4$ and $S_12=246$

(iv)  given $a_3$  = 15,$S_{10}$ = 125, find $d$ and $a_{10}$  . 
    Solution:
    Since $S_n=n/2(a+l)$	 
          where $a$=first term and $l$=last term.	
      $S_10=10/2(a+a_10)$	  
      $125=5(a+a_10)$	  
      $25=(a+a_10)................(1)$	
   
    Since $S_n=n/2(2a+(n-1)d)$
	      $125=10/2(2a+9d)$
	      $25=(2a+9d)...............(2)$
	
	Also given,$a_3$  = 15
    Therefore  $a_3=a+(3-1)d$	
			$15=a+2d ...............(3)$
	  
    Solving equation (2) and (3)
	Multiplying eq (3) by 2 and subracting with equation (2)
    $-5=5d$	
   i.e; $d=-1$	
	Puttng the value of $d$ in equation (3) and find $a$
	$15=a+2(-1)$
	$a=17$
	
	Putting the value of $a=17$ in equation (1) and find $a_10$
	$a+a_10=25$
	$17+a_10=25$
	i.e; $a_10=8$
	
Ans:$d=-1$ and $a_10=8$


(v)  given $d$ = 5, $S_9$  = 75, find $a$ and $a_9$.


(vi)  given $a$ = 2,  $d$ = 8,  Sn = 90, find $n$ and $a_n$. 


(vii)  given $a$ = 8, $a_n$ = 62, Sn = 210, find $n$ and $d$. 


(viii)  given $a_n$ = 4, $d$ = 2,  Sn = –14, find $n$ and $a$.


(ix)  given $a$ = 3, $n$ = 8, S = 192, find $d$.


(x)  given $l$ = 28,  S = 144,  and there are total 9 terms. Find $a$.