The sum of the third and the seventh terms of a AP is 6 and their product is 8 find the sum of sixteen terms of the A.P.
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A3*a7=8
from equation 1 we get
a7=6-a3
=A3(6-a3)=8
=(a3×a3)-6a3+8=0. Sony square not working
=(a3×a3)-2a3-4a3+8=0
A3(a3-2)and -4(a3-2)=0
so A3 can be 2 or 4
if A3=2 , then a7 =4
if A3=4 ,then a7=2
from equation 1 we get
a7=6-a3
=A3(6-a3)=8
=(a3×a3)-6a3+8=0. Sony square not working
=(a3×a3)-2a3-4a3+8=0
A3(a3-2)and -4(a3-2)=0
so A3 can be 2 or 4
if A3=2 , then a7 =4
if A3=4 ,then a7=2
Pranathi0306:
It is sorry not sony. Sorry for the mistake
Answered by
0
the third term is a3=a+(3-1)d=a+2d, the seventh term is a7=a+(7-1)d=a+6d, a+2d+a+6d=6,2a+8d=6,2(a+4d)=6,a+4d=3 equation (1) ,a=3-4d(put this value in Equation (2), (a+2d)(a+6d)=8, a square +6ad+2ad+12d square=8,a square +8ad+12d square=8 this is equation (2), (3-4d) square+8ad+12d square=8,9+16d square -24d square +8ad+12ad square
sorry for mistake of half answer but i can solve this question and answer is 94, i can send the solution in a few minutes in the last line in my soln. replace 12ad by 12 d square
please mark it as a brainiest answer
next step
8ad+28d square-24d square -1=o
8ad+4d square -1=0
d(8a+4d)-1=0,8a+4d-1=0 equation 3, solving equation (1) and (3), a+4d-3=0 and 8a+4d-1=0 we can subtract, we get a=-2/7 we put this value in a= 3-4d then we find d=23/28 then we find sum of 16 th term of an A.P. be by using the formula of sum that is n/2+2a+(n-1)d, the n is 16, we can solve this we find S16=94 answer
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