Question

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

Answer 1

See explanation.

Explanation:

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

Given:

`n=19`

`a_19=103`

`S_19=1102`

To calculate:

`a_1`, `a_2`, `a_3`

First we can use the sum formula to calculate `a_1`:

`S_19=(a_1+a_19)/2*19`

`1102=(a_1+103)/2*19`

`2204=(a_1+103)*19`

`a_1+103=116`

`a_1=13`

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

`a_1+18*d=a_19`

`13+18*d=103`

`18*d=90`

`d=5`

Now having `a_1` and `d` we can calculate `a_2` and `a_3`:

`a_2=a_1+d=13+5=18`

`a_3=a_2+d=18+5=23`

Answer:

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

Answer 2

`13,18,23.`

Explanation:

Here,

`n=19, a_n=103 , S_n=1102`.

We know that,

`n^(th) term` of Arithmetic series is`=a_n`

and sum of first n-terms`=S_n`

Now, `a=`first term and

`d=`common ratio of Arithmetic series.

So,

`color(red)(a_n=a+(n-1)d and S_n=n/2[2a+(n-1)d]`

`=>103=a+(19-1)d and 1102=19/2[2a+(19-1)d]`

`=>103=a+18d and1102xx2/19=2a+18d`

`=>103=a+18d and 116=2a+18d`

Let,

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

From `(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

`" "`
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`