Question

The sum of the third and
Seventh term of an AP and
their product is 8. Find the
Sum of first 16 terms of the
Ap​

Answer 1

Answer:

The third and seventh term of the ap can be written as a+2d and a+6d respectively.

a+2d +a+6d = 6

2a + 8d = 6

a+4d = 3

So, a = 3-4d

Now, (a+2d)(a+6d) = 8 (given)

we can also write as , (3-4d + 2d)(3-4d+6d) = 8 (a=3-4d as derived above)

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

$9 + 6d - 6d + {4d}^{2} = 8$

$9 + {4d}^{2} = 8$

${4d}^{2} = 8 - 9$

$4 {d}^{2} = - 1$

${d}^{2} = \frac{ - 1}{4}$

$d = \sqrt{ \frac{ -1}{4} }$

$d = \frac{1}{2}$

Now we have to find the sixteenth term, i.e a+15d

$a = 3-4d \: and \: (d)= \frac{1}{2}$

$So, a=3 - 4 \times \frac{1}{2} </p><p></p><p>$

a = 1

$a _{16} = a + 15d$

$1 + 15 \times \frac{1}{2}$

$\frac{17}{2}$

If it is correct , then thumbs up please!!!!!!!!!!!!!!!!!!!!!!!

Answer 2

Answer:

17/2

Step-by-step explanation:

The third and seventh term of the ap can be written as a+2d and a+6d respectively.

a+2d +a+6d = 6

2a + 8d = 6

a+4d = 3

So, a = 3-4d

Now, (a+2d)(a+6d) = 8 (given)

we can also write as , (3-4d + 2d)(3-4d+6d) = 8 (a=3-4d as derived above)

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

9 + 6d - 6d + {4d}^{2} = 89+6d−6d+4d

2

=8

9 + {4d}^{2} = 89+4d

2

=8

{4d}^{2} = 8 - 94d

2

=8−9

4 {d}^{2} = - 14d

2

=−1

{d}^{2} = \frac{ - 1}{4}d

2

=

4

−1

d = \sqrt{ \frac{ -1}{4} }d=

4

−1

d = \frac{1}{2}d=

2

1

Now we have to find the sixteenth term, i.e a+15d

a = 3-4d \: and \: (d)= \frac{1}{2}a=3−4dand(d)=

2

1

So, a=3 - 4 \times \frac{1}{2}So,a=3−4×

2

1

a = 1

a _{16} = a + 15da

16

=a+15d

1 + 15 \times \frac{1}{2}1+15×

2

1

\frac{17}{2}

2

17