Question

If the sum of 3 and 7 term of AP is 6 and their product is 8. Find the sum of first 30 terms

Answer 1

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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 * Seventh term = (a+2d)*(a+6d) = 8 
(3-4d+2d)*(3-4d+6d) = 8==> (3-2d)*(3+2d) = 8 
i.e. 9-4d^2 = 8==> d^2 = (9-8)/4 = 0.25==> d = 0.5 or -0.5 

Now to check which is correct d... 
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*0.5 = 2 
7th term = a+6d= 1+6*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*(-0.5) = 4 
7th term = a+6d= 5+6*(-0.5) = 2 
Sum = 6 and Product = 8 

Since both are matching, we will go with bothvalues

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

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

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

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