Question

Show work and explain with formulas.

4. If a_6 = 18 and a_13 = 32, find the sum of the 13 terms in an arithmetic sequence.

5. Find the sum (see picture).

6. Find the first three terms of the arithmetic series: a_1 = 17, a_n = 197, S_n = 2247
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Answer 1

`\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}`

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=(a_1+n_1)/(2)\cdot n`

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

`S_(13)=(8+32)/(2)\cdot13=(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=(2a_1+(n-1)d)/(2)\cdot n`

Substitute n = 7, a₁ = 7 and d = 2:

`S_7=(2(7)+(7-1)(2))/(2)\cdot7=(14+(6)(2))/(2)\cdot7=(14+12)/(2)\cdot7=(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=(2a_1+(n-1)d)/(2)\cdot n`

Substitute:

`(1)\qquad 197=17+(n-1)d\\\\(2)\qquad2247=((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=(34+180)/(2)\cdot n\\\\2247=(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`