Question

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

Answer 1

Sum of third and seventh terms = 6.

a + 2d + a + 6d = 6

2a + 8d = 6

a + 4d = 3

a = 3 - 4d  ---- (i)

Here also given: product is 8.

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

(3 - 4d + 2d) * (3 - 4d + 6d) = 8

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

9 - 4d^2 = 8

d = (1/2)

Apply in (i), we get

a = 3 - 4(1/2)

  = 1

Sum of first n terms : (n/2)[2a + (n - 1) * d]

Sum of first 16 terms = (16/2)[2 + (16 - 1) * 1/2]

                                  = 8[2 + 15/2]

                                  = 76

#BeBrainly.

Answer 2

Answer :-

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

3^2 – 2d^2 = 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

Hope it's Helpful.....:)