Question

the sum of the third and the seventh term of an ap is 6 and their product is 8 find the sum of the first 16 term of an ap​

Answer 1

$\large \bold{ \underline{ \underline{ \: Answer : \: \: \: }}}$

$\to$ The sum of first 16 terms is 76

$\large \bold{ \underline{ \underline{ \: Explaination : \: \: \: }}}$

Given ,

The sum of the third and the seventh term of the given AP is 6

$\star$ a₃ + a₇ = 6

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

$\to$ 2a + 8d = 6

$\to$ 2( a + 4d ) = 6

$\to$ a + 4d = 3

$\to$ a = 3 - 4d ---------- (i)

Also , the product of third term and seventh term is 8

$\star$ a₃ × a₇ = 8

$\to$ ( a + 2d ) × ( a + 6d ) = 8

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

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

$\to$ ( 3 )² - (2d )² = 8

$\to$ 9 - 4d² = 8

$\to$ 4d² = 9 - 8

$\to$ 4d² = 1

$\to$ d² = 1/4

$\to$ d = 1/2

Put the value of d = 1/2 in eq (i)

$\to$ a = 3 - 4 × 1/2

$\to$ a = 3 - 2

$\to$ a = 1

Thus , sum of 16th term is :

$\implies \bold{S_{16} } = \frac{n}{2} \{2a + (n - 1)d \} \\ \\ \implies \bold{S_{16}} = \frac{16}{2} \{ 2 \times 1 + (16 - 1) \frac{1}{2} \} \\ \\ \implies \bold{S_{16}} = 8 \{2 + \frac{15}{2} \} \\ \\ \implies \bold{S_{16} } =8\bigg( 2 + 7.5 \bigg)\\ \\ \implies \bold{S_{16}} =8 \times 9.5 \\ \\ \implies \bold{S_{16} } = 76$

$\therefore$

The sum of first 16 terms is 76