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.

Maths

Class 10

821

Michael

Qayanat

Answer / Solution

Answer / Solution

**Solution:**

From the given statements, a_{3 }+_{ }a_{7} = 6 .....(i) and a_{3 }×_{ }a_{7 }= 8 ........(ii),

By using the n^{th} term formula, a_{n} = a + (n − 1)d, the third term will be

Third term, a_{3 }= a + (3 − 1)d ⇒ a_{3 }= a + 2d ………(iii) and

seventh term, a_{7} = a + (7 -1)d ⇒ a_{7 }= a + 6d ....... (iv)

Putting in equation (i) from the values from equation (iii) and (iv), we get,

a + 2d + a + 6d ⇒ 6 2a + 8d = 6 ⇒ a + 4d = 3

⇒ 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 – 4d^{2} = 8

4d^{2} = 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 = 1/2

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

We know, the sum of nth term of AP is; S_{n} = n/2 [2a + (n – 1)d], So, when a = 1 and d = 1/2. Then, the sum of first 16 terms are;

S_{16} = 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;

S_{16} = 16/2 [2(5)+ (16 – 1)(–1/2)] = 8(5/2) = 20

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