Question

the sum of 3rd and 7th term of an ap is 6 and product is 8 find the sum of first 16 terms of this AP​

Answer 1

Answer:

285/4 ,75/4

Step-by-step explanation:Given a3 + a7 = 6, a3*a7=8

a3= a+2d,

a7= a+6d ,

a3+a7=a+2d+a+6d=2a+8d=6

a+4d=3  ⇒a=3-4d

(a+2d)(a+6d)=8

substituting value of a = 3-4d

(3-2d)(3+2d)=8

9-4d^2=8  ⇒d= +1/2,-1/2

Then a=1,5.

so S16= 15/2(2a+15d)

S16 = 285/4,75/4

Answer 2

Answer:

the terms are

a-4d,a-3d,a-2d,a-d,a,a+d,a+2d,a+3d

3rd term=a-2d

7th term=a+2d

a3+a7=a-2d+a+2d=6

2a=6

a=2/6

a=3

(a-2d) (a+2d)=8

a2-4d2=8

4d2=a2-8

4d2=3 2-8

4d2=9-8

4d2=1

d2=1 l / 2

d=+-1/2

taking d =1/2

s16=16/2{2(a-4d)+(16-1)d}

8 [2(3-4×1/2)+15×1/2]

8[2+15/2]

=8×19/2 =76

taking d=-1/2

s16 =16/2[2(a-4d)+(16-1)d

8=[2(3-4×-1/2)+(15×-1/2)

8=20-15/2=8×5/2

=20

s16= 20and76