Question

The sum of first terms of an AP is given by Sn=2n2 +8n . Find the sixteenth term of the AP. ​

Answer 1

EXPLANATION.

Sum of first terms of an A.P.

⇒ Sₙ = 2n² + 8n.

As we know that,

⇒ Tₙ = Sₙ - Sₙ₋₁.

⇒ 2n² + 8n - [2(n - 1)² + 8(n - 1)].

⇒ 2n² + 8n - [2(n² + 1 - 2n) + 8n - 8].

⇒ 2n² + 8n - [2n² + 2 - 4n + 8n - 8].

⇒ 2n² + 8n - [2n² + 4n - 6].

⇒ 2n² + 8n - 2n² - 4n + 6.

⇒ 8n - 4n + 6.

⇒ 4n + 6. = Algebraic expression.

As we know that,

Put the value of n = 1 in equation, we get.

⇒ 4(1) + 6.

⇒ 4 + 6.

⇒ 10.

Put the value of n = 2 in equation, we get.

⇒ 4(2) +  6.

⇒ 8 + 6.

⇒ 14.

Put the value of n = 3 in equation, we get.

⇒ 4(3) + 6.

⇒ 12 + 6.

⇒ 18.

Put the value of n = 4 in equation, we get.

⇒ 4(4) + 6.

⇒ 16 + 6.

⇒ 22.

Their Series = 10, 14, 18, 22,,,,,,,

First term of an A.P. = a = 10.

Common difference = d = b - a = 14 - 10 = 4.

As we know that,

General terms of an A.P.

⇒ Tₙ = a + (n - 1)d.

⇒ T₁₆ = a + (16 - 1)d.

⇒ T₁₆ = a + 15d.

⇒ T₁₆ = 10 + 15(4).

⇒ T₁₆ = 10 + 60.

⇒ T₁₆ = 70.

                                                                                                                         

MORE INFORMATION.

Supposition of terms in A.P.

(1) = Three terms as : a - d, a, a + d.

(2) = Four terms as : a - 3d, a - d, a + d, a + 3d.

(3) = Five terms as : a - 2d, a - d, a, a + d, a + 2d.

Answer 2

Answer :-

• Given :-

The sum of first n terms of an Ap is Sn = 2n²+8n

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• To Find ::

the sixteenth term of the Ap.

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Solution :-

Let us consider the Sum of first Term be n = 1

Since substituting the values of n in the given equation.

• S1 = 2 × 1²+ 8 × 1 = 10
• S2 = 2 × 2²+8×2 = 24
• S3 = 2 × 3²+8×3 = 42

We obtained the Sum of S = 1 2 and 3.

• S1 = 10 which is also lthe first term of the Ap = a.

• Second term of the Ap a2 = a + d = a2 = 10 + d

By simple Assuming Method We can see that.

• The S2 Sum of two terms = 24 We know the first term = 10 and another number which is the second term is added to obtain sum of 24.

• S2 = 24 = 10+ a2

• 24 - 10 = a2

• Hence second term of Ap = 14.

Therefore, a2 - a1 = d

• 14-10 = 4

• d ( common difference ) = 14.

Calculation of 16th term of the Ap.

$\sf{\pink{ \star {an = a + (n - 1)d}{} }{} }{}$

$\sf{a16 = 10 + (16 - 1)4}{} \\ \\ = 10 + 15 \times 4 \\ \\ a16 = 10 + 60 \\ \\ \sf{ \orange{a16 = 70}{} }{}$

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• Hence the 16th term of the Ap = 70.