Question

The sum of the third and the seventh term of an A.P is 6 and their product is 8. Find the sum of the first 16 terms of the AP.

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Answer 1

Answer:

76 (or) 20

Step-by-step explanation:

Let a and d be the first term and common difference of A.P.

We know that nth term of an AP a(n) = a + (n - 1) * d.

(i) Sum of an AP:

Sum of the third and the seventh tern of an AP is 6.

a₃ + a₇ = 6.

⇒ [a + (3 - 1) * d] + (a + (7 - 1) * d] = 6

⇒ a + 2d + a + 6d = 6

⇒ 2a + 8d = 6

⇒ a + 4d = 3

⇒ a = 3 - 4d

(ii) Product of an AP:

a₃ * a₇ = 8

⇒ [a + (3 - 1) * d] [a + (7 - 1) * d] = 8

⇒ (a + 2d)(a + 6d) = 8

⇒ (3 - 4d + 2d)(3 - 4d + 6d) = 8

⇒ (3 - 2d)(3 - 2d) = 8

⇒ 9 - 4d² = 8

⇒ 4d² = 1

⇒ d² = 1/4

⇒ d = ±1/2

When a = 1, d = (1/2):

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

                                  = (16/2)[2 + (16 -1) * 1/2]

                                  = 8[2 + 15/2]

                                  = 76.

When a = 5, d = -1/2:

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

                                  = (16/2)[10 + (16 - 1) * -1/2]

                                  = 8[10 - 15/2]

                                  = 20.

Therefore, Sum of first 16 terms of the AP = 76, 20.

Hope it helps!

Answer 2

$\bf\huge\boxed{\boxed{\bf\huge\:Hello\:Mate}}}$

$\bf\huge Let\: the\: AP\: be\: a - 4d , a - 3d , a - 2d , a - d , a , a + d , a + 2d , a + 3 d$

$\bf\huge => a_{3} = a - 2d$

$\bf\huge => a_{7} = a - 2d$

$\bf\huge => a_{3} + a_{7} = a - 2d + a - 2d = 6$

$\bf\huge => 2a = 6$

$\bf\huge => a = 3 (Eqn 1)$

$\bf\huge Hence\: (a - 2d) (a + 2d) = 8$

$\bf\huge => a^2 - 4d^2 = 8$

$\bf\huge => 4d^2 = a^2 - 8$

$\bf\huge => 4d^2 = (3)^2 - 8 = 9 - 8 = 1$

$\bf\huge => d^2 = \frac{1}{4}$

$\bf\huge => d = \frac{1}{2}$

$\bf\huge\texttt Hence$

$\bf\huge S_{16} = \frac{16}{2} [2\times (a - 4d)+ (16 - 1)\times d]$

$\bf\huge => 8[2\times (3 - 4\times \frac{1}{2})+ 15\times \frac{1}{2}]$

$\bf\huge => 8[2 + \frac{15}{2}]= 8\times \frac{19}{2} = 76$

$\bf\huge => d = - \frac{1}{2}$

$\bf\huge Putting\:the\: Value\: of\: D$

$\bf\huge S_{16} = \frac{16}{2} [2\times (a - 4d)+ (16 - 1)\times d]$

$\bf\huge => 8[2\times (3 - 4\times - \frac{1}{2})+ 15\times - \frac{1}{2}]$

$\bf\huge => 8[2\times 5 - \frac{15}{2}]$

$\bf\huge => 8 [ \frac{20 - 15}{2}]$

$\bf\huge => 8\times \frac{5}{2} = 20$

$\bf\huge Hence$

$\bf\huge S_{16} = 20 , 76$

$\bf\huge\boxed{\boxed{\:Regards=\:Yash\:Raj}}}$