Question

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Answer 1

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 terms of the AP.

Answer:

The sum of the first sixteen terms of the AP is either 76 or 20.

Step-by-step-explanation:

Let the first term of the AP be "a" and the common difference be "d".

We know that,

tₙ = a + ( n - 1 ) * d

From the first condition,

t₃ + t₇ = 6

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

⇒ a + 2d + a + 6d = 6

⇒ a + a + 2d + 6d = 6

⇒ 2a + 8d = 6

⇒ a + 4d = 3 - - - [ Dividing by 2 ]

⇒ a = 3 - 4d

⇒ a = - 4d + 3 - - - ( 1 )

From the second condition,

t₃ * t₇ = 8

⇒ ( a + 2d ) ( a + 6d ) = 8

⇒ ( - 4d + 3 + 2d ) ( - 4d + 3 + 6d ) = 8 - - - [ From ( 1 ) ]

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

⇒ ( - 2d + 3 ) ( 2d + 3 ) = 8

⇒ ( 3 - 2d ) ( 3 + 2d ) = 8

⇒ ( 3 )² - ( 2d )² = 8 - - - [ ( a + b ) ( a - b ) = a² - b² ]

⇒ 9 - 4d² = 8

⇒ 9 - 8 = 4d²

⇒ 4d² = 1

⇒ d² = 1 / 4

⇒ d = ± ( 1 / 2 )

⇒ d = ± 0.5

Now, we know that,

Sₙ = ( n / 2 ) [ 2a + ( n - 1 ) * d ]

⇒ S₁₆ = ( 16 / 2 ) [ 2 * ( - 4d + 3 ) + ( 16 - 1 ) * d ]

⇒ S₁₆ = 8 ( - 8d + 6 + 15d )

By using d = 0.5, we get,

⇒ S₁₆ = 8 ( - 8 * 0.5 + 6 + 15 * 0.5 )

⇒ S₁₆ = 8 ( - 4 + 6 + 7.5 )

⇒ S₁₆ = 8 * ( 2 + 7.5 )

⇒ S₁₆ = 8 * 9.5

⇒ S₁₆ = 76

Now, by using d = - 0.5, we get,

⇒ S₁₆ = 8 [ - 8 * ( - 0.5 ) + 6 + 15 * ( - 0.5 ) ]

⇒ S₁₆ = 8 ( 8 * 0.5 + 6 - 15 * 0.5 )

⇒ S₁₆ = 8 ( 4 + 6 - 7.5 )

⇒ S₁₆ = 8 * ( 10 - 7.5 )

⇒ S₁₆ = 8 * 2.5

⇒ S₁₆ = 20

∴ The sum of the first sixteen terms of the AP is either 76 or 20.

Answer 2

Answer:

The sum of the third and the seventh tems of an AP is 6 and their products is 8. Find the sum of first sixteen terms of the AP.

so using formula for n th term

$\sf \: a_n=[a+(n-1)d]$

A= first term

N= no. of terms

D= common difference

$\gray{\underbrace{\huge\underline{ \red{\sf Given:-}}}}$

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The third term of an A.P is given by,

Third term = a₁ + 2d

Seventh term of the A.P is given by,

Seventh term = a₁ + 6d

By first case given,

a₁ + 2d + a₁ + 6d = 6

2a1 + 8d = 6

Divide the whole equation by 2

a1 + 4d = 3

• a₁ = 3 - 4d-------(1)

Now by second case given,

• (a₁ + 2d) × (a₁ + 6d) = 8

Substitute the value of a₁ from equation 1

• (3 - 4d + 2d) × (3 - 4d + 6d) = 8

• (3 - 2d) × (3 + 2d) = 8

By using the identity,

(a + b) × (a - b) = a² - b²,

• 9 - 4d² = 8

• -4d² = -1

• d² = 1/4

• d = ± 1/2

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Case 1:-

• If d = 1/2

Substitute the value of d in equation 1,

A₁ = 3 - 4 × 1/2

• A₁ = 3 - 2
• A₁ = 1

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Now sum of n terms of an A.P is given by,

$\:\\ \sf\: Now, \: sum \: of \: first \: sixteen \: terms \: of \: the \: AP\\ \\ \sf\: s_{16}\: = \: \frac{16}{2} \: [2a + (16 - 1) \: d] \: \: \: \: \bigg[ \because \: s_n \: = \: \frac{n}{2} [2a + (n - 1) \: d] \bigg]\\ \\ \sf{S_{16}=\dfrac{16}{2}\bigg(2\times 1+(16-1)\times \frac{1}{2} \bigg)}\\ \sf S₁₆ = 8 (2 + \frac{15}{2})\\ \sf S₁₆ = 8 (2 + 7.5)\\S₁₆ = 8 × 9.5\\ \sf S₁₆ = 76\\ \\ \sf \: When, \: d \: = - \frac{1}{2} \: then \: from \: Eq. \: (i) ,\: we \: get \\ \\ \sf When,d=− \bf \: a + 4 \bigg(- \frac{1}{2} \bigg) = 3 \implies \: a - = 3 \implies \: a = 5 \\ \\ \sf \: Now, \: sum \: of \: first \: sixteen \: terms \: of \: this \: AP, \\ \\ \sf\: s_{16} \: = \: \frac{16}{2} \: [2a + (16 - 1) \: d] \: \: \: \: \: \: \: \bigg [ \because \: \: s_n \: = \: \frac{n} {2} \: [2a + (n - 1) \: d] \bigg ] \\\\ \sf\: = 8 \: (2a + 15d)=8(2a+15d)\\ \\ \sf \: = 8 \bigg [2(5) + 15 \bigg( - \frac{1}{2} \bigg) \bigg] = 8 \bigg(10 - \frac{15}{2} \bigg) = 8 \bigg( \frac{5}{2} \bigg) = 20 \\ \\ \red{\underline{\underline{\sf \:\purple{ Hence, \: sum \: of \: first \: sixteen \: terms, \: s_{16} \: \bold{20} \: or \: \bold{76}}}}}$

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Substitute the data,