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.

Answer 1

Let the first term be a and common difference be d

nth term = a+(n-1)d

Given Third Term + Seventh term = (a+2d)+(a+6d) = 6 ==> a+4d = 3

hence, a= 3-4d

Third Term * Seventh term = (a+2d)*(a+6d) = 8

(3-4d+2d)*(3-4d+6d) = 8==> (3-2d)*(3+2d) = 8

i.e. 9-4d^2 = 8==> d^2 = (9-8)/4 = 0.25==> d = 0.5 or -0.5

Now to check which is correct d...

Substitute and find

Case (a): d= 0.5

a+4d = 3==> a=3-4d = 3-4(0.5)=1

3rd term = a+2d= 1+2*0.5 = 2

7th term = a+6d= 1+6*0.5 = 4

Sum = 6 and Product = 8

Case (b): d= -0.5

a+4d = 3==> a=3-4d = 3-4(-0.5) = 3+2 = 5

3rd term = a+2d= 5+2*(-0.5) = 4

7th term = a+6d= 5+6*(-0.5) = 2

Sum = 6 and Product = 8

Since both are matching, we will go with bothvalues

Sum of first 16 terms = n*(2a+(n-1)d)/2 = 16*(2a+15d)/2

= 8*(2a+15d)

Case (a): d= 0.5

Sum = 8*(2*1+15*0.5)=76

Case (b): d= 0.5

Sum = 8*(2*5+15*(-0.5))=20

Answer 2

▶ 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.


▶ 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,

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


→ Similarly, when d = -1/2 .

Then,

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


✔✔ Hence, it is solved ✅✅.


$\huge \green{ \boxed{ \boxed{ \mathscr{THANKS}}}}$