Question

The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP.​

Answer 1

$\huge\underline\mathrm{Question}:-$

The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP.

$\huge\underline\mathrm{Solution}:-$

From the given statements, we can write,

a3 + a7 = 6 ___________(i)

And

a3 × a7 = 8 ___________(ii)

By the nth term formula,

an = a + (n − 1)d

Third term, a3 = a + (3 -1)d

a3 = a + 2d___________(iii)

And Seventh term, a7 = a + (7 -1)d

a7 = a + 6d _________(iv)

From equation (iii) and (iv), putting in equation(i), we get,

a + 2d + a + 6d = 6

2a + 8d = 6

a+4d=3

or

a = 3 – 4d __________(v)

Again putting the eq. (iii) and (iv), in eq. (ii), we get,

(a + 2d) × (a + 6d) = 8

Putting the value of a from equation (v), we get,

(3 – 4d + 2d) × (3 – 4d + 6d) = 8

(3 – 2d) × (3 + 2d) = 8

32 – 2d2 = 8

9 – 4d2 = 8

4d2 = 1

d = 1/2 or -1/2

Now, by putting both the values of d, we get,

a = 3 – 4d = 3 – 4(1/2) = 3 – 2 = 1, when d = ½

a = 3 – 4d = 3 – 4(-1/2) = 3+2 = 5, when d = -1/2

We know, the sum of nth term of AP is;

Sn = n/2 [2a + (n – 1)d]

So, when a = 1 and d=1/2

Then, the sum of first 16 terms are;

S16 = 16/2 [2 + (16 – 1)1/2] = 8(2+15/2) = 76

And when a = 5 and d= -1/2

Then, the sum of first 16 terms are;

S16 = 16/2 [2.5+ (16 – 1)(-1/2)] = 8(5/2)=20

Sum of the first sixteen terms of the AP is 20.

Answer 2

A N S W E R :

• The sum of the first sixteen terms of the AP is 76

Given :

• Sum of the third and the seventh terms of an AP is 6 and their products is 8

To find :

• Find the sum of the first sixteen terms of the AP ?

Solution :

• a_3 + a_7 = 6

=> a + 2d + a + 6d = 6

=> 2a + 8d = 6

=> a + 4d = 3

=> a = 3 - 4d (equation 1)

Also, Given that,

• a_3 × a_7 = 8

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

Putting the equation value of (1) & (2),

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

=> 9 - 4d² = 8

=> -4d² = -1

=> d = 1/2

Putting the value of d in (1),

=> a = 3 - 4(1/2)

=> a = 3 - 2

=> a = 1

As we know that,

• Sn = n/2[2a + (n - 1)d]

We have to calculate the values, S16,

=> S16 = 16/2[2(1) + (15)(1/2)]

=> S16 = 8(2 + 15/2)

=> S16 = 8(19/2)

=> S16 = 4 × 19

=> S16 = 76

Hence,

• The sum of the sixteen terms of the AP is 76.