As numbers are in A.P. 
Thus (y + 10) - 2y = (3y + 2) - (y + 10) 
=> 10 - y = 2y - 8 
=> -3y = -18 
=> y = 6  

205x15= (200+5) x15
=200x15+5x15  

 For 3Y-2X to be minimum the condition is that Y must be substituted with least value and X must be with large value 
=> 3(3)-2(3)
=3.  

 ((991+771)2-(991-771)2)/(991x771)=((A+B)2-(A-B)2)/(AxB) 
(4xAB)/AB=4  

 L.C.M. of 9 and 6 = 18
∴ required numbers are 108, 126, 144, 162, 180, 198 which are 6. 

 Required sum = 100 + 102 + ... + 200 which is an A.P. where a = 100, d = 2, l = 200. 
Using formula Tn = a + (n - 1)d 
Tn = 100 + (n-1)2 = 200 
=> 2n = 200 - 98 = 102 
=> n = 51 
Now Using formula Sn = (n/2)(a + l) 
∴ Required sum = (51/2)(100+200)  = 51 x 150  = 7550 

  
 Here numbers are in G.P.  Here a = 5,  r = 2, l = 1280. 
 Using formula Tn = arn- 1 
 Tn = 5 x 2(n-1) = 1280  
 =2(n-1) = 256  
 =2(n-1) 
 = 28  
 => n - 1 = 8  => n = 9 

   
 let the ages be 7 , 7.25, 7.5 and so on  
 Here a = 7, d = 1/4 , Sn = 250   
 Using formula Sn = (n/2)[2a + (n-1)d]   
 => (n/2)[14+(n-1)(1/4)]  = 250  
 => n[14 + (n-1)/4] = 500  
 => n[56 + (n-1)] = 2000  
 => n[n + 55] = 2000  
 => n2 + 55n - 2000 = 0  
 => n2 + 80n -25n - 2000 = 0  
 => n(n-80) -25(n-80) = 0  
 => (n-80)(n-25) = 0  
 => n = 25 

 Let's have first term as a, common difference is d then 
a + 3d = 16 ... (i) 
a + 11d = 80 ... (ii) 
Subtracting (i) from (ii) 
=> 8d = 64  => d = 8 
Using (i) 
=> a = 14 - 3d  = -10  

 Let's have first term as a, common difference is d then 
a + 3d = 16 ... (i) 
a + 11d = 80 ... (ii) 
Subtracting (i) from (ii) 
=> 8d = 64 
=> d = 8 
Using (i) 
=> a = 14 - 3d  = -10
Using formula Tn = a + (n - 1)d   
T17 = -10 + (17 - 1) x 8
= 118