Question

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

Answer 1

▶ 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 .