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THE SUM OF THE 3RD AND 7TH TERMS OF AN AP IS 6 AND THEIR PRODUCT IS 8 FIND THE SUM OF 1ST 20 TERMSOF AP
Let the first term be a and common difference be d 
nth term = a + (n - 1) d 

Given Third Term + Seventh term = (a + 2d) + (a + 6d) = 6 ==> a + 4d = 3 
hence, a= 3 - 4d 

Third Term x Seventh term = (a + 2d) (a+6d) = 8 
(3 - 4d + 2d) (3 - 4d + 6d) = 8==> (3 - 2d)(3 + 2d) = 8 
i.e. 9 - 4d2 = 8 ==> d2 = (9 - 8)/4 = 0.25==> d = 0.5 or -0.5 

Substitute and find 

Case (a): d = 0.5 
a + 4d = 3==> a = 3 - 4d = 3 - 4(0.5) =1 
3rd term = a + 2d= 1+2 x 0.5 = 2 
7th term = a + 6d= 1+6 x 0.5 = 4 
Sum = 6 and Product = 8 

Case (b): d= -0.5 
a + 4d = 3==> a=3-4d = 3-4(-0.5) = 3 + 2 = 5 
3rd term = a + 2d= 5+2 x (-0.5) = 4 
7th term = a + 6d= 5+6 x (-0.5) = 2 
Sum = 6 and Product = 8 

Since both are matching, we will go with both values.

Now,  

Sum of first 16 terms = n x (2a + (n - 1)d)/2 = 16 x (2a + 15d) /2  
= 8 x (2a + 15d) 

Case (a): d= 0.5 
Sum = 8x (2 x 1+15 x 0.5) = 76 

Case (b): d= 0.5 
Sum = 8 x (2 x 5+15 x (-0.5)) = 20


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