Answer:
[tex]\large\boxed{4.\ S_{13}=260}[/tex]
[tex]\large\boxed{5.\ \sum\limits_{k=1}^7(2k+5)=91}[/tex]
[tex]\large\boxed{6.\ a_1=17,\ a_2=26,\ a_3=35}[/tex]
Step-by-step explanation:
4.
We have:
[tex]a_6=18,\ a_{13}=32[/tex]
These are the terms of the arithmetic sequence.
We know:
[tex]a_n=a_1+(n-1)d[/tex]
Therefore
[tex]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[/tex]
Substitute a₆ = 18 and a₁₃ = 32:
[tex]7d=32-18[/tex]
[tex]7d=14[/tex] divide both sides by 7
[tex]d=2[/tex]
[tex]a_6=a_1+5d[/tex]
[tex]18=a_1+5(2)[/tex]
[tex]18=a_1+10[/tex] subtract 10 from both sides
[tex]8=a_1\to a_1=8[/tex]
The formula of a sum of terms of an arithmetic sequence:
[tex]S_n=\dfrac{a_1+n_1}{2}\cdot n[/tex]
Substitute a₁ = 8, a₁₃ = 32 and n = 13:
[tex]S_{13}=\dfrac{8+32}{2}\cdot13=\dfrac{40}{2}\cdot13=20\cdot13=260[/tex]
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5.
We have
[tex]\sum\limits_{k=3}^7(2k+5)\to a_k=2k+5[/tex]
Calculate [tex]a_{k+1}[/tex]
[tex]a_{k+1}=2(k+1)+5=2k+2+5=2k+7[/tex]
Calculate the difference:
[tex]a_{k+1}-a_k=(2k+7)-(2k+5)=2k+7-2k-5=2[/tex]
It's the arithmetic sequence with first term
[tex]a_1=2(1)+5=2+5=7[/tex]
and common difference d = 2.
The formula of a sum of terms of an arithmetic sequence:
[tex]S_n=\dfrac{2a_1+(n-1)d}{2}\cdot n[/tex]
Substitute n = 7, a₁ = 7 and d = 2:
[tex]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[/tex]
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6.
We have:
[tex]a_1=17,\ a_n=197,\ S_n=2247[/tex]
The formula for the n-th term of an arithmetic sequence:
[tex]a_n=a_1+(n-1)d[/tex]
The formula of the sum of terms of an arithmetic sequence:
[tex]S_n=\dfrac{2a_1+(n-1)d}{2}\cdot n[/tex]
Substitute:
[tex](1)\qquad 197=17+(n-1)d\\\\(2)\qquad2247=\dfrac{(2)(17)+(n-1)d}{2}\cdot n[/tex]
Convert the first equation:
[tex]197=17+(n-1)d[/tex] subtract 17 from both sides
[tex]180=(n-1)d[/tex] Substitute it to the second equation:
[tex]2247=\dfrac{34+180}{2}\cdot n\\\\2247=\dfrac{214}{2}\cdot n[/tex]
[tex]2247=107n[/tex] divde both sides by 107
[tex]21=n\to n=21[/tex]
Put the value of n to the equation (n - 1)d = 180:
[tex](21-1)d=180[/tex]
[tex]20d=180[/tex] divide both sides by 20
[tex]d=9[/tex]
Therefore we have the explicit formula for the nth term of an arithmetic sequence:
[tex]a_n=17+(n-1)(9)\\\\a_n=17+9n-9\\\\a_n=9n+8[/tex]
Put n = 1, n = 2 and n = 3:
[tex]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[/tex]
