How do you find the first three terms of the arithmetic series n=19, $a_{n} = 103$ $a_n=103$ , $S_{n} = 1102$ $S_n=1102$ ?

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How do you find the first three terms of the arithmetic series n=19, $a_{n} = 103$ $a_n=103$ , $S_{n} = 1102$ $S_n=1102$ ?

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Answer 1 · 2018-05-03T09:24:06

See explanation.

Explanation:

First we have to write everything which is given and what we are looking for:

Given:

$n = 19$ $n=19$

$a_{19} = 103$ $a_19=103$

$S_{19} = 1102$ $S_19=1102$

To calculate:

$a_{1}$ $a_1$ , $a_{2}$ $a_2$ , $a_{3}$ $a_3$

First we can use the sum formula to calculate $a_{1}$ $a_1$ :

$S_{19} = \frac{a_{1} + a_{19}}{2} \cdot 19$ $S_19=(a_1+a_19)/2*19$

$1102 = \frac{a_{1} + 103}{2} \cdot 19$ $1102=(a_1+103)/2*19$

$2204 = (a_{1} + 103) \cdot 19$ $2204=(a_1+103)*19$

$a_{1} + 103 = 116$ $a_1+103=116$

$a_{1} = 13$ $a_1=13$

Now we can calculate the common difference using two given terms:

$a_{1} + 18 \cdot d = a_{19}$ $a_1+18*d=a_19$

$13 + 18 \cdot d = 103$ $13+18*d=103$

$18 \cdot d = 90$ $18*d=90$

$d = 5$ $d=5$

Now having $a_{1}$ $a_1$ and $d$ $d$ we can calculate $a_{2}$ $a_2$ and $a_{3}$ $a_3$ :

$a_{2} = a_{1} + d = 13 + 5 = 18$ $a_2=a_1+d=13+5=18$

$a_{3} = a_{2} + d = 18 + 5 = 23$ $a_3=a_2+d=18+5=23$

Answer:

The first three terms are: $13, 18$ $13,18$ and $23$ $23$ .

Answer 2 · 2018-05-03T09:29:20

$13, 18, 23 .$ $13,18,23.$

Explanation:

Here,

$n = 19, a_{n} = 103, S_{n} = 1102$ $n=19, a_n=103 , S_n=1102$ .

We know that,

$n^{t h} t e r m$ $n^(th) term$ of Arithmetic series is $= a_{n}$ $=a_n$

and sum of first n-terms $= S_{n}$ $=S_n$

Now, $a =$ $a=$ first term and

$d =$ $d=$ common ratio of Arithmetic series.

So,

$a_{n} = a + (n - 1) d and S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ $color(red)(a_n=a+(n-1)d and S_n=n/2[2a+(n-1)d]$

$\Rightarrow 103 = a + (19 - 1) d and 1102 = \frac{19}{2} [2 a + (19 - 1) d]$ $=>103=a+(19-1)d and 1102=19/2[2a+(19-1)d]$

$\Rightarrow 103 = a + 18 d and 1102 \times \frac{2}{19} = 2 a + 18 d$ $=>103=a+18d and1102xx2/19=2a+18d$

$\Rightarrow 103 = a + 18 d and 116 = 2 a + 18 d$ $=>103=a+18d and 116=2a+18d$

Let,

$a + 18 d = 103 \to (1) and 2 a + 18 d = 116 \to (2)$ $color(blue)(a+18d=103 to(1)and 2a+18d=116to(2)$

From $(2)$ $(2)$ we get,

$a+color(blue)(a+18d)=116...to[2a=a+a]$

$a+color(blue)(103)=116...tocolor(blue)([use (1) ]$

$=>a=116-103$

$=>color(violet)(a=13$

From $(1)$ we have,

$13+18d=103$

$=>18d=103-13$

$=>18d=90$

$=>color(violet)(d=5$

Hence, first three terms of Arithmetic series are :

$(i)a=color(brown)(13$

$(ii)a+d=13+5=color(brown)(18$

$(iii)a+2d=13+2xx5=13+10=color(brown)(23$

Answer 3 · 2018-05-03T14:57:37

$" "$
**The first three terms: ** $color(blue)(a_1 = 13, a_2=18 and a_3=23$

Explanation:

$" "$
**Total number of terms: ** $color(red)(n=19$

**19th term: ** $color(red)(a_19=103$

**Sum of the first 19 terms: ** $color(red)(S_19=1102$

In an Arithmetic Sequence, ** the difference between one term and the next is a Common Difference**: $color(red)d$ .

The terms are:

$color(brown)(a_1, (a_1+d), (a_1 + 2d), (a_1+3d), ... ,$ where $color(brown)(a_1$ is the First Term and $color(brown)(d$ is the Common Difference.

The Sum of an arithmetic sequence is called an Arithmetic Series.

$color(green)("Step 1:"$

$color(blue)("Formula 1:"$

Sum to $color(red)(n)$ terms of an arithmetic series:

$color(green)(S_n = [n(a_1+a_n)]/2$ .

$color(blue)("Formula 2:"$

**Find the ** $color(red)(n^(th) term$ of the arithmetic series:

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

$color(green)("Step 2:"$

Since, $color(red)(n, a_n, and S_n)$ are given, use $color(blue)("Formula 1"$ to find $color(red)(a_1)$ .

$1102=[19(a_1+a_(19))]/2$

$1102=[19(a_1+103)]/2$

$1102=(19a_1+1957)/2$

Multiply both sides of the equation by $color(red)(2$ .

$1102*color(red)(2) =(19a_1+1957)/cancel 2*color(red)(cancel 2$

$2204=19a_1+1957$

Flipping sides:

$19a_1+1957=2204$

Subtract $color(red)(1957$ .

$19a_1+cancel 1957-color(red)(cancel 1957)=2204-color(red)(1957)$

$19a_1=247$

Divide both sides by $color(red)(19$

$(cancel 19a_1)/color(red)(cancel 19)=247/color(red)(19$

$a_1=247/19=13$

$:. "First Term "= a_1 = 13$

$color(green)("Step 3:"$

Use $color(blue)("Formula 2"$ to find the Common Difference $color(red)(d)$ .

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

When $n=19, a_n = a_19$

$103=13+d(19-1)$

$103=13+19d-d$

$103=13+18d$

Flipping sides:

$13+18d=103$

Subtract $color(red)13$ .

$cancel 13+18d-color(red)cancel 13=103-color(red)13$

$18d=90$

Divide by $color(red)(18$ .

$(cancel 18d)/color(red)(cancel 18)=cancel 90^color(green)(5)/color(red)(cancel 18$

$:. "Common Difference " color(red)(d=5$

Terms: $13+[13+1(5)]+[13+2(5)]$

So, $a_1 = 13, a_2=18 and a_3=23$

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How do you find the first three terms of the arithmetic series n=19, $a_{n} = 103$ $a_n=103$ , $S_{n} = 1102$ $S_n=1102$ ?

3 Answers

Explanation:

$S_{19} = \frac{a_{1} + a_{19}}{2} \cdot 19$ $S_19=(a_1+a_19)/2*19$

$1102 = \frac{a_{1} + 103}{2} \cdot 19$ $1102=(a_1+103)/2*19$

$2204 = (a_{1} + 103) \cdot 19$ $2204=(a_1+103)*19$

$a_{1} + 103 = 116$ $a_1+103=116$

$a_{1} = 13$ $a_1=13$

$a_{1} + 18 \cdot d = a_{19}$ $a_1+18*d=a_19$

$13 + 18 \cdot d = 103$ $13+18*d=103$

$18 \cdot d = 90$ $18*d=90$

$d = 5$ $d=5$

Explanation:

Explanation:

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How do you find the first three terms of the arithmetic series n=19, a_n=103an=103a_n=103, S_n=1102Sn=1102S_n=1102?

3 Answers

Explanation:

S_19=(a_1+a_19)/2*19S19=a1+a192⋅19S_19=(a_1+a_19)/2*19

1102=(a_1+103)/2*191102=a1+1032⋅191102=(a_1+103)/2*19

2204=(a_1+103)*192204=(a1+103)⋅192204=(a_1+103)*19

a_1+103=116a1+103=116a_1+103=116

a_1=13a1=13a_1=13

a_1+18*d=a_19a1+18⋅d=a19a_1+18*d=a_19

13+18*d=10313+18⋅d=10313+18*d=103

18*d=9018⋅d=9018*d=90

d=5d=5d=5

Explanation:

Explanation:

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How do you find the first three terms of the arithmetic series n=19, $a_{n} = 103$ $a_n=103$ , $S_{n} = 1102$ $S_n=1102$ ?

$S_{19} = \frac{a_{1} + a_{19}}{2} \cdot 19$ $S_19=(a_1+a_19)/2*19$

$1102 = \frac{a_{1} + 103}{2} \cdot 19$ $1102=(a_1+103)/2*19$

$2204 = (a_{1} + 103) \cdot 19$ $2204=(a_1+103)*19$

$a_{1} + 103 = 116$ $a_1+103=116$

$a_{1} = 13$ $a_1=13$

$a_{1} + 18 \cdot d = a_{19}$ $a_1+18*d=a_19$

$13 + 18 \cdot d = 103$ $13+18*d=103$

$18 \cdot d = 90$ $18*d=90$

$d = 5$ $d=5$