Question

Show work and explain with formulas.

Answer 1

Answer:

$\large\boxed{4.\ S_{13}=260}$

$\large\boxed{5.\ \sum\limits_{k=1}^7(2k+5)=91}$

$\large\boxed{6.\ a_1=17,\ a_2=26,\ a_3=35}$

Step-by-step explanation:

4.

We have:

$a_6=18,\ a_{13}=32$

These are the terms of the arithmetic sequence.

We know:

$a_n=a_1+(n-1)d$

Therefore

$a_6=a_1+(6-1)d\ \text{and}\ a_{13}=a_1+(13-1)d\\\\a_6=a_1+5d\ \text{and}\ a_{13}=a_1+12d\\\\a_{13}-a_6=(a_1+12d)-(a_1+5d)\\\\a_{13}-a_6=a_1+12d-a_1-5d\\\\a_{13}-a_6=7d$

Substitute a₆ = 18 and a₁₃ = 32:

$7d=32-18$

$7d=14$             divide both sides by 7

$d=2$

$a_6=a_1+5d$

$18=a_1+5(2)$

$18=a_1+10$            subtract 10 from both sides

$8=a_1\to a_1=8$

The formula of a sum of terms of an arithmetic sequence:

$S_n=\dfrac{a_1+n_1}{2}\cdot n$

Substitute a₁ = 8, a₁₃ = 32 and n = 13:

$S_{13}=\dfrac{8+32}{2}\cdot13=\dfrac{40}{2}\cdot13=20\cdot13=260$

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5.

We have

$\sum\limits_{k=3}^7(2k+5)\to a_k=2k+5$

Calculate $a_{k+1}$

$a_{k+1}=2(k+1)+5=2k+2+5=2k+7$

Calculate the difference:

$a_{k+1}-a_k=(2k+7)-(2k+5)=2k+7-2k-5=2$

It's the arithmetic sequence with first term

$a_1=2(1)+5=2+5=7$

and common difference d = 2.

The formula of a sum of terms of an arithmetic sequence:

$S_n=\dfrac{2a_1+(n-1)d}{2}\cdot n$

Substitute n = 7, a₁ = 7 and d = 2:

$S_7=\dfrac{2(7)+(7-1)(2)}{2}\cdot7=\dfrac{14+(6)(2)}{2}\cdot7=\dfrac{14+12}{2}\cdot7=\dfrac{26}{2}\cdot7=(13)(7)\\\\S_7=91$

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6.

We have:

$a_1=17,\ a_n=197,\ S_n=2247$

The formula for the n-th term of an arithmetic sequence:

$a_n=a_1+(n-1)d$

The formula of the sum of terms of an arithmetic sequence:

$S_n=\dfrac{2a_1+(n-1)d}{2}\cdot n$

Substitute:

$(1)\qquad 197=17+(n-1)d\\\\(2)\qquad2247=\dfrac{(2)(17)+(n-1)d}{2}\cdot n$

Convert the first equation:

$197=17+(n-1)d$          subtract 17 from both sides

$180=(n-1)d$    Substitute it to the second equation:

$2247=\dfrac{34+180}{2}\cdot n\\\\2247=\dfrac{214}{2}\cdot n$

$2247=107n$             divde both sides by 107

$21=n\to n=21$

Put the value of n to the equation (n - 1)d = 180:

$(21-1)d=180$

$20d=180$              divide both sides by 20

$d=9$

Therefore we have the explicit formula for the nth term of an arithmetic sequence:

$a_n=17+(n-1)(9)\\\\a_n=17+9n-9\\\\a_n=9n+8$

Put n = 1, n = 2 and n = 3:

$a_1=9(1)+8=9+8=17\qquad\text{CORRECT :)}\\\\a_2=9(2)+8=18+8=26\\\\a_3=9(3)+8=27+8=35$