Question

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

Answer 1

Answer:

so using formula for n th term 

a_n=a+(n-1)dan=a+(n−1)d

a = first term

n = no. of terms 

d = common difference

So given

a + (3-1)d + a (7-1)d = 6

2a + 8d = 6

a + 4d = 3

a = 3 - 4d

(a + 2d)(a + 6d) = 8 

(3 - 2d)( 3 + 2d) = 8   {substituting a = 3 - 2d}

9 - 4d² = 8

d = +1/2 and - 1/2

so when d = +1/2 then a = 3 - 2 = 1 and when d = -1/2  a = 3 + 2 = 5

so using formula for 

S_n=\frac{n}{2}[2a+(n-1)d]Sn=2n[2a+(n−1)d]

when d = +1/2

S_n=\frac{16}{2}[2*1+(16-1)d]Sn=216[2∗1+(16−1)d]

S_n=76Sn=76

Similarly when d = -1/2

S_n=20Sn=20

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.....:)