Math / Algebra

Question8. Find the indicated arithmetic term. a) $a=5, d=3$ ; find $t_{12}$ c) $a=-\frac{3}{4}, d=\frac{1}{2}$ ; find $t_{10}$ e) $a=-0.75, d=0.05$ ; find $t_{40}$

Studdy Solution

STEP 1

What is this asking? We're looking for specific terms in arithmetic sequences, given the starting term and the common difference. Watch out! Don't mix up the starting term and the common difference!
Also, remember that the term number starts at 1, not 0!

STEP 2

1. Find $t_{12}$
2. Find $t_{10}$
3. Find $t_{40}$

STEP 3

Alright, let's kick things off with the formula for the $n$ th term of an arithmetic sequence: $t_n = a + (n-1) \cdot d$ .
Remember, $a$ is our initial value, $d$ is our common difference, and $n$ is the term number we're looking for.

STEP 4

We're given $a = 5$ , $d = 3$ , and we want to find $t_{12}$ , so $n = 12$ .
Let's plug these values into our formula!

STEP 5

$t_{12} = 5 + (12-1) \cdot 3$

STEP 6

$t_{12} = 5 + (11) \cdot 3$

STEP 7

$t_{12} = 5 + 33$

STEP 8

$t_{12} = 38$ So, the 12th term is 38!

STEP 9

Let's use our formula again: $t_n = a + (n-1) \cdot d$ .
This time, $a = -\frac{3}{4}$ , $d = \frac{1}{2}$ , and we want $t_{10}$ , so $n = 10$ .

STEP 10

Substituting gives us: $t_{10} = -\frac{3}{4} + (10-1) \cdot \frac{1}{2}$

STEP 11

$t_{10} = -\frac{3}{4} + (9) \cdot \frac{1}{2}$

STEP 12

$t_{10} = -\frac{3}{4} + \frac{9}{2}$

STEP 13

To add these fractions, we need a common denominator, which is 4.
So we multiply the second fraction by $\frac{2}{2}$ (which is equal to one, so we're not changing the value): $t_{10} = -\frac{3}{4} + \frac{9 \cdot 2}{2 \cdot 2}$

STEP 14

$t_{10} = -\frac{3}{4} + \frac{18}{4}$

STEP 15

$t_{10} = \frac{-3 + 18}{4}$

STEP 16

$t_{10} = \frac{15}{4}$ Awesome! The 10th term is $\frac{15}{4}$ !

STEP 17

One more time with the formula: $t_n = a + (n-1) \cdot d$ .
We have $a = -0.75$ , $d = 0.05$ , and we're looking for $t_{40}$ , so $n = 40$ .

STEP 18

Let's plug in the values: $t_{40} = -0.75 + (40-1) \cdot 0.05$

STEP 19

$t_{40} = -0.75 + (39) \cdot 0.05$

STEP 20

$t_{40} = -0.75 + 1.95$

STEP 21

$t_{40} = 1.2$ The 40th term is 1.2!

STEP 22

a) $t_{12} = 38$ c) $t_{10} = \frac{15}{4}$ e) $t_{40} = 1.2$

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Question8. Find the indicated arithmetic term. a) a=5,d=3a=5, d=3a=5,d=3; find t12t_{12}t12​ c) a=−34,d=12a=-\frac{3}{4}, d=\frac{1}{2}a=−43​,d=21​; find t10t_{10}t10​ e) a=−0.75,d=0.05a=-0.75, d=0.05a=−0.75,d=0.05; find t40t_{40}t40​

Studdy Solution

STEP 1

What is this asking? We're looking for specific terms in arithmetic sequences, given the starting term and the common difference. Watch out! Don't mix up the starting term and the common difference!Also, remember that the term number starts at 1, not 0!

STEP 2

1. Find t12t_{12}t12​2. Find t10t_{10}t10​3. Find t40t_{40}t40​

STEP 3

Alright, let's **kick things off** with the formula for the nnnth term of an arithmetic sequence: tn=a+(n−1)⋅dt_n = a + (n-1) \cdot dtn​=a+(n−1)⋅d.Remember, aaa is our **initial value**, ddd is our **common difference**, and nnn is the term number we're looking for.

STEP 4

We're given a=5a = 5a=5, d=3d = 3d=3, and we want to find t12t_{12}t12​, so n=12n = 12n=12.Let's **plug these values** into our formula!

STEP 5

t12=5+(12−1)⋅3t_{12} = 5 + (12-1) \cdot 3t12​=5+(12−1)⋅3

STEP 6

t12=5+(11)⋅3t_{12} = 5 + (11) \cdot 3t12​=5+(11)⋅3

STEP 7

t12=5+33t_{12} = 5 + 33t12​=5+33

STEP 8

t12=38t_{12} = 38t12​=38 So, the **12th term** is 38!

STEP 9

Let's **use our formula** again: tn=a+(n−1)⋅dt_n = a + (n-1) \cdot dtn​=a+(n−1)⋅d.This time, a=−34a = -\frac{3}{4}a=−43​, d=12d = \frac{1}{2}d=21​, and we want t10t_{10}t10​, so n=10n = 10n=10.

STEP 10

**Substituting** gives us: t10=−34+(10−1)⋅12t_{10} = -\frac{3}{4} + (10-1) \cdot \frac{1}{2}t10​=−43​+(10−1)⋅21​

STEP 11

t10=−34+(9)⋅12t_{10} = -\frac{3}{4} + (9) \cdot \frac{1}{2}t10​=−43​+(9)⋅21​

STEP 12

t10=−34+92t_{10} = -\frac{3}{4} + \frac{9}{2}t10​=−43​+29​

STEP 13

To **add these fractions**, we need a common denominator, which is 4.So we multiply the second fraction by 22\frac{2}{2}22​ (which is equal to one, so we're not changing the value): t10=−34+9⋅22⋅2t_{10} = -\frac{3}{4} + \frac{9 \cdot 2}{2 \cdot 2}t10​=−43​+2⋅29⋅2​

STEP 14

t10=−34+184t_{10} = -\frac{3}{4} + \frac{18}{4}t10​=−43​+418​

STEP 15

t10=−3+184t_{10} = \frac{-3 + 18}{4}t10​=4−3+18​

STEP 16

t10=154t_{10} = \frac{15}{4}t10​=415​ Awesome! The **10th term** is 154\frac{15}{4}415​!

STEP 17

One more time with the formula: tn=a+(n−1)⋅dt_n = a + (n-1) \cdot dtn​=a+(n−1)⋅d.We have a=−0.75a = -0.75a=−0.75, d=0.05d = 0.05d=0.05, and we're looking for t40t_{40}t40​, so n=40n = 40n=40.

STEP 18

Let's **plug in the values**: t40=−0.75+(40−1)⋅0.05t_{40} = -0.75 + (40-1) \cdot 0.05t40​=−0.75+(40−1)⋅0.05

STEP 19

t40=−0.75+(39)⋅0.05t_{40} = -0.75 + (39) \cdot 0.05t40​=−0.75+(39)⋅0.05

STEP 20

t40=−0.75+1.95t_{40} = -0.75 + 1.95t40​=−0.75+1.95

STEP 21

t40=1.2t_{40} = 1.2t40​=1.2 The **40th term** is 1.2!

STEP 22

a) t12=38t_{12} = 38t12​=38 c) t10=154t_{10} = \frac{15}{4}t10​=415​ e) t40=1.2t_{40} = 1.2t40​=1.2

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Question8. Find the indicated arithmetic term. a) $a=5, d=3$ ; find $t_{12}$ c) $a=-\frac{3}{4}, d=\frac{1}{2}$ ; find $t_{10}$ e) $a=-0.75, d=0.05$ ; find $t_{40}$

What is this asking? We're looking for specific terms in arithmetic sequences, given the starting term and the common difference. Watch out! Don't mix up the starting term and the common difference!
Also, remember that the term number starts at 1, not 0!

1. Find $t_{12}$
2. Find $t_{10}$
3. Find $t_{40}$

Alright, let's kick things off with the formula for the $n$ th term of an arithmetic sequence: $t_n = a + (n-1) \cdot d$ .
Remember, $a$ is our initial value, $d$ is our common difference, and $n$ is the term number we're looking for.

We're given $a = 5$ , $d = 3$ , and we want to find $t_{12}$ , so $n = 12$ .
Let's plug these values into our formula!

$t_{12} = 5 + (12-1) \cdot 3$

$t_{12} = 5 + (11) \cdot 3$

$t_{12} = 5 + 33$

$t_{12} = 38$ So, the 12th term is 38!

Let's use our formula again: $t_n = a + (n-1) \cdot d$ .
This time, $a = -\frac{3}{4}$ , $d = \frac{1}{2}$ , and we want $t_{10}$ , so $n = 10$ .

Substituting gives us: $t_{10} = -\frac{3}{4} + (10-1) \cdot \frac{1}{2}$

$t_{10} = -\frac{3}{4} + (9) \cdot \frac{1}{2}$

$t_{10} = -\frac{3}{4} + \frac{9}{2}$

To add these fractions, we need a common denominator, which is 4.
So we multiply the second fraction by $\frac{2}{2}$ (which is equal to one, so we're not changing the value): $t_{10} = -\frac{3}{4} + \frac{9 \cdot 2}{2 \cdot 2}$

$t_{10} = -\frac{3}{4} + \frac{18}{4}$

$t_{10} = \frac{-3 + 18}{4}$

$t_{10} = \frac{15}{4}$ Awesome! The 10th term is $\frac{15}{4}$ !

One more time with the formula: $t_n = a + (n-1) \cdot d$ .
We have $a = -0.75$ , $d = 0.05$ , and we're looking for $t_{40}$ , so $n = 40$ .

Let's plug in the values: $t_{40} = -0.75 + (40-1) \cdot 0.05$

$t_{40} = -0.75 + (39) \cdot 0.05$

$t_{40} = -0.75 + 1.95$

$t_{40} = 1.2$ The 40th term is 1.2!

a) $t_{12} = 38$ c) $t_{10} = \frac{15}{4}$ e) $t_{40} = 1.2$