(2) The sum of third and seventh term of an A. P. is 6 and their produ

1.	(2) The sum of third and seventh term of an A. P. is 6 and their product is8. Find the first term and the common difference of the A. P
Answer» Let the first term be a and common difference be dnth term = a+(n-1)d Given Third Term + Seventh term = (a+2d)+(a+6d) = 6 ==> a+4d = 3hence, a= 3-4d Third Term * Seventh term = (a+2d)(a+6d) = 8(3-4d+2d)(3-4d+6d) = 8==> (3-2d)(3+2d) = 8i.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= 0.5a+4d = 3==> a=3-4d = 3-4(0.5)=13rd term = a+2d= 1+20.5 = 27th term = a+6d= 1+60.5 = 4Sum = 6 and Product = 8 Case (b): d= -0.5a+4d = 3==> a=3-4d = 3-4(-0.5) = 3+2 = 53rd term = a+2d= 5+2(-0.5) = 47th term = a+6d= 5+6(-0.5) = 2Sum = 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= 0.5Sum = 8(21+150.5)=76 Case (b): d= 0.5Sum = 8(25+15(-0.5))=20

(2) The sum of third and seventh term of an A. P. is 6 and their product is8. Find the first term and the common difference of the A. P

Answer»

Let the first term be a and common difference be dnth term = a+(n-1)d

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

Third Term * Seventh term = (a+2d)*(a+6d) = 8(3-4d+2d)*(3-4d+6d) = 8==> (3-2d)*(3+2d) = 8i.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= 0.5a+4d = 3==> a=3-4d = 3-4(0.5)=13rd term = a+2d= 1+2*0.5 = 27th term = a+6d= 1+6*0.5 = 4Sum = 6 and Product = 8

Case (b): d= -0.5a+4d = 3==> a=3-4d = 3-4(-0.5) = 3+2 = 53rd term = a+2d= 5+2*(-0.5) = 47th term = a+6d= 5+6*(-0.5) = 2Sum = 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= 0.5Sum = 8*(2*1+15*0.5)=76

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