Question

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Do solve this Question.

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


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Answer 1

Given:

$\implies$ a₃ + a₇ = 6

$\implies$ a₃ × a₇ = 8

We have:

$\implies$ a₃ + a₇ = 6

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

$\implies$ 2a + 8d = 6

Now:

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

$\implies$ a + 4d = 6/2

$\implies$ a + 4d = 3

Therefore:

$\huge\boxed{\sf{a = 3 - 4d}}$ - - - - (1)

And:

$\implies$ a₃ × a₇ = 8

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

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

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

Now:

$\implies$ 3² - (2d)² = 8

$\implies$ 9 - 4d² = 8

$\implies$ 4d² = 9 - 8

$\implies$ 4d² = 1

$\implies$ d² = 1/4

$\implies$ d = √(1/4)

Therefore: $\boxed{\sf{d = \frac{1}{2}, d= - \frac{1}{2}}}$

Putting the value of d in equation (1),

We get:

$\implies$ a = 3 - 4 × 1/2

$\implies$ a = 3 - 2

Therefore: a = 1

Using this formula:

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

Sum of 16th terms:

$\implies$ $\sf{S_{16} = \frac{n}{2} [2a + (n - 1)d] }$

$\implies$ $\sf{\frac{16}{2} [2 \times 1 + (16 - 1) \frac{1}{2}] }$

$\implies$ $\sf{8(2 + \frac{15}{2} )}$

$\implies$ $\sf{8(2 + 7.5)}$

$\implies$ $\sf{8(9.5)}$

$\implies$ $\sf{76}$

Or:

$\implies$ $\sf{S_{16} = \frac{n}{2} [2a + (n - 1)d] }$

$\implies$ $\sf{\frac{16}{2} [2 \times 5 + (16 - 1) \frac{-1}{2}] }$

$\implies$ $\sf{8(10 - 15 \times -\frac{1}{2} )}$

$\implies$ $\sf{8(-5 \times -\frac{1}{2})}$

$\implies$ $\sf{8(\frac{5}{2})}$

$\implies$ $\sf{8(2.5)}$

$\implies$ $\sf{20}$

Therefore:

The sum of first 16 terms of the AP is 76 or 20.

Answer 2

▶ Answer :-

→ 76 and 20 .

▶ Step-by-step explanation :-

➡ Given :-

→ $a_3 + a_7 = 6 .$

→$a_3 \times a_7 = 8 .$

➡ To find :-

→ The sum of first 16th term of an AP .

$\huge \pink{ \mid \underline{ \overline{ \tt Solution :- }} \mid}$

Here,

a = first term

n = no. of terms

d = common difference

We have,

∵ $a_3 + a_7 = 6 .$

⇒a + ( 3 - 1 )d + a( 7 - 1 )d = 6 .

⇒2a + 8d = 6.

⇒a + 4d = 3 .

∴ a = 3 - 4d .

▶ And,

∵ $a_3 \times a_7 = 8 .$

⇒(a + 2d)(a + 6d) = 8 .

⇒(3 - 2d)( 3 + 2d) = 8 { putting a = 3 - 2d, we get }

⇒9 - 4d² = 8 .

⇒d = ± 1/2 . [ +1/2 and - 1/2 ]

So, when d = +1/2 then a = 3 - 2 = 1 and when d = -1/2, then a = 3 + 2 = 5 .

▶ So, using formula

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

→ When, d = +1/2 .

Then,

$\begin{lgathered}\sf \because S_{16} = \frac{16}{2} \bigg(2 \times 1 + (16 - 1) \frac{1}{2} \bigg). \\ \\ \sf = \frac{16}{2} \bigg(2 + \frac{15}{2} \bigg). \\ \\ = \sf \frac{16}{2} \bigg( \frac{4 + 15}{2} \bigg). \\ \\ \sf = 8 \times \frac{19}{2} . \\ \\ \huge \orange{ \boxed{ \boxed{ \sf \therefore S_{16} = 76.}}}\end{lgathered}$

→ Similarly, when d = -1/2 .

Then,

$\begin{lgathered}\sf \because S_{16} = \frac{16}{2} \bigg(2 \times5 + (16 - 1) \frac{ - 1}{2} \bigg). \\ \\ \sf = \frac{16}{2} \bigg(10 - \frac{15}{2} \bigg). \\ \\ = \sf \frac{16}{2} \bigg( \frac{20 - 15}{2} \bigg). \\ \\ \sf = 8 \times \frac{5}{2} . \\ \\ \huge \orange{ \boxed{ \boxed{ \sf \therefore S_{16} = 20.}}}\end{lgathered}$

Hence, it is solved .