Question

In an a3+a7=6 and a3×a7=8 is given. Find the sum of first 16 terms of an AP.​

Answer 1

Given:

In an A.P. a3 + a7=6 and a3 × a7=8  

To find:

The sum of the first 16 terms of an A.P.

Solution:

We know the formula of the nth term of an A.P. is:

$\boxed{\bold{a_n = a + (n-1)d}}$

 where a = first term, n = no. of terms and d = common difference

So,

$a_3 + a_7 = 6$

$\implies a + (3-1)d + a + (7-1)d = 6$

$\implies a + 2d + a + 6d = 6$

$\implies 2a + 8d = 6$

$\implies a + 4d = 3$

$\implies a = 3 - 4d$ ↔ equation1

 

$a_3 \times a _7 =8$

$\implies (a + 2d)(a + 6d) = 8$

on substituting the value of a from equation1, we get

$\implies (3 - 4d + 2d)(3 - 4d + 6d) = 8$

$\implies (3 - 2d )(3 + 2d) = 8$

since a² - b² = (a + b)(a - b)

$\implies 9 - 4d^2 = 8$

$\implies 4d^2 = 1$

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

$\implies d =$ ± $\frac{1}{2}$

 

When d = $+\frac{1}{2}$ ⇔ Then a = $3 - 4\times \frac{1}{2}$ = $3 - 2 = 1$

When d = $-\frac{1}{2}$ ⇔ Then a = $3 - 4\times -\frac{1}{2}$ = $3 + 2 = 5$

 

We know the formula of the sum of n terms of an A.P. is:

$\boxed{\bold{S_n = \frac{n}{2}[2a + (n - 1)d] }}$

Therefore,  

The sum of the first 16 terms of the given A.P. will be:

When a = 1 and d = $+\frac{1}{2}$:

$S_1_6 = \frac{16}{2}[2 + (16 - 1)(\frac{1}{2} )] = \boxed{\bold{76}}$

and

When a = 5 and d = $-\frac{1}{2}$:

$S_1_6 = \frac{16}{2}[(2\times 5) + (16 - 1)(-\frac{1}{2} )] = \boxed{\bold{20}}$

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