Question

The sum of the third and the seventh terms of an Ap is 6 and their product is 8.find the sum of first sixteen term of ap.

Answer 1

so using formula for n th term 
$a_n=a+(n-1)d$

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

So given

a + (3-1)d + a (7-1)d = 6
2a + 8d = 6
a + 4d = 3
a = 3 - 4d

(a + 2d)(a + 6d) = 8 
(3 - 2d)( 3 + 2d) = 8   {substituting a = 3 - 2d}

9 - 4d² = 8

d = +1/2 and - 1/2

so when d = +1/2 then a = 3 - 2 = 1 and when d = -1/2  a = 3 + 2 = 5

so using formula for 
$S_n=\frac{n}{2}[2a+(n-1)d]$

when d = +1/2

$S_n=\frac{16}{2}[2*1+(16-1)d]$

$S_n=76$

Similarly when d = -1/2

$S_n=20$

Answer 2

Given :-

→ a₃ + a₇ = 6.

→ a₃ × a₇ = 8 .

To find :-

→ S₁₅ .

Solution :-

We have,

→ a₃ + a₇ = 6.

⇒ a + 2d + a + 6d = 6 .

⇒ 2a + 8d = 6 .

⇒ 2( a + 4d ) = 6 .

⇒ a + 4d = 6/2 .

⇒ a + 4d = 3

∵a = 3 - 4d ............(1) .

And,

→ a₃ × a₇ = 8 .

⇒ ( a + 2d ) × ( a + 6d ) = 8 .

⇒ ( 3 - 4d + 2d ) ( 3 - 4d + 6d ) = 8 .

⇒ ( 3 - 2d )( 3 + 2d ) = 8 .

⇒ 3² - (2d)² = 8 .

⇒ 9 - 4d² = 8.

⇒ 4d² = 9 - 8 .

⇒ 4d² = 1 .

⇒ d² = 1/4 .

⇒ d = √(1/4) .

∴ d = 1/2 .

Putting the value of d in equation (1), we get ,

⇒ a = 3 - 4 × 1/2 .

⇒ a = 3 - 2 .

∴ a = 1 .

Thus, sum of 16th term is given by ,

$\begin{lgathered}\because \sf S_{16} = \frac{n}{2} \{2a + (n - 1)d \}. \\ \\ \sf = \frac{16}{2} \{ 2 \times 1 + (16 - 1) \frac{1}{2} \}. \\ \\ \sf = 8 \{2 + \frac{15}{2} . \} \\ \\ \sf= 8 \bigg( 2 + 7.5 \bigg). \\ \\ \sf = 8 \times 9.5 . \\ \\ \huge \pink{ \boxed{ \tt\therefore S_{16} = 76 .}}\end{lgathered}$

Hence, sum of first 16 terms is 76.