Question

the sum of third and seventh terms of a ap is 6 and their product is 8 find the sum of 1​

Answer 1

Correct Question :-

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

Solution :-

► a₃ + a₇ = 6

⇒ a + 2d + a + 6d = 6

⇒ 2a + 8d = 6

⇒ a + 4d = 3

⇒ a = 3 - 4d                                    ...(i)

a₃ × a₇ = 8⇒(a + 2d) × (a + 6d) = 8

⇒ (3 - 4d + 2d) × (3 - 4d + 6d) = 8           ...[from eq.(i)]

(3 - 2d)(3 + 2d) = 8

⇒ (3)² - (2d)² = 8

⇒ 9 - 4d² = 8

⇒ -4d² = 8 - 9

⇒ -4d² = -1

⇒ 1 = 4d²

⇒ d = √(1/4)

⇒ d = 1/2

___________________

Putting the value of d = 1/2 in eq.(i) :-

a = 3 - 4d

⇒ a = 3 - (4*1/2)

⇒ a = 3 - 2

⇒ a = 1

___________________

Now, sum of the first 16 terms :-

$\displaystyle{S_1_6}=\frac{n}{2} [2a + (n-1)d]\\\\\displaystyle{\implies S_1_6=\frac{16}{2}[2*1+(16-1)\frac{1}{2} }\\\\\displaystyle{\implies S_1_6=8(2+\frac{15}{2}) }\\\\\displaystyle{\implies S_1_6=16 + 60}\\\\ \huge \yellow{ \boxed{ \tt\therefore S_{16} =76}}$

Answer 2

Answer:

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

Solution :-

► a₃ + a₇ = 6

⇒ a + 2d + a + 6d = 6

⇒ 2a + 8d = 6

⇒ a + 4d = 3

⇒ a = 3 - 4d ...(i)

a₃ × a₇ = 8⇒(a + 2d) × (a + 6d) = 8

⇒ (3 - 4d + 2d) × (3 - 4d + 6d) = 8 ...[from eq.(i)]

(3 - 2d)(3 + 2d) = 8

⇒ (3)² - (2d)² = 8

⇒ 9 - 4d² = 8

⇒ -4d² = 8 - 9

⇒ -4d² = -1

⇒ 1 = 4d²

⇒ d = √(1/4)

⇒ d = 1/2

___________________

Putting the value of d = 1/2 in eq.(i) :-

a = 3 - 4d

⇒ a = 3 - (4*1/2)

⇒ a = 3 - 2

⇒ a = 1

___________________

Now, sum of the first 16 terms :-

\begin{gathered}\displaystyle{S_1_6}=\frac{n}{2} [2a + (n-1)d]\\\\\displaystyle{\implies S_1_6=\frac{16}{2}[2*1+(16-1)\frac{1}{2} }\\\\\displaystyle{\implies S_1_6=8(2+\frac{15}{2}) }\\\\\displaystyle{\implies S_1_6=16 + 60}\\\\ \huge \yellow{ \boxed{ \tt\therefore S_{16} =76}}\end{gathered}

Step-by-step explanation:

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