Question

The sum of 3 rd nd 7 th terms of an A.P. is 6 nd there product is 8 .Find the sum of first 20 terms of the A.P

Answer 1

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

Answer 2

Let the first term be 'a' and the common difference be 'd'
Given : a₃ + a₇ = 6 and
            a₃*a₇ = 8 (or) a₃*(6-a₃) = 8 ⇒ (a₃)² - 6a₃ + 8 = 0
            The roots for the above equation can be obtained by splitting the middle term or using the quadratic equation. The roots are 2 and 4. 
a₃ = 2 or 4 and a₇ = 4 or 2 (from given conditions) 
So,  Case-1:  a + 2d = 2 and a + 6d = 4 . On subtracting both the equations we get d = 1/2 and a = 3 
Thus, S₂₀ = (20/2)[ 2(1) + (20-1)(1/2)] = 115.
Case-2 : a + 2d = 4 and a + 6d = 2. On solving, we get d = -1/2 and a = 5.
Thus, S₂₀ = (20/2)[ 2(5) + (20-1)(-1/2)] = 5 

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