Question

show that the sequence 9,12,15,18 an ap find its 16th and 24 term​

Answer 1

$\bold{\underline{\underline{\huge{\sf{AnsWer:}}}}}$

The sequence is an AP.

16th term of the AP = 54

24th term of the AP = 78

$\bold{\underline{\underline{\large{\sf{StEp\:by\:stEp\:explanation:}}}}}$

GiVeN :

• Sequence :

1. 9,12,15,18.....

To FiNd :

• Whether the sequence is an AP
• The 16th term of the AP
• The 24th term of the AP

SoLuTiOn :

We have,

• $\bold{\sf{t_1}}$ = a = 9
• $\bold{\sf{t_2}}$ = 12
• $\bold{\sf{t_3}}$ = 15
• $\bold{\sf{t_4}}$ = 18

So, to prove that the sequence is in AP we have to calculate the common difference 'd' and if the common difference is constant for all the given terms, we say that the sequence is an AP.

Let's begin!

Common difference, d = $\bold{\sf{t_2}}$ - $\bold{\sf{t_1}}$

Block in the values,

$\hookrightarrow$ $\sf{12\:-\:9}$

$\hookrightarrow$ $\sf{3}$

•°• The common difference for the first two terms is 3.

Let's calculate the common difference for the next two terms.

$\hookrightarrow$ $\bold{\sf{t_3}}$ - $\bold{\sf{t_2}}$

$\hookrightarrow$ $\sf{15\:-\:12}$

$\hookrightarrow$ $\sf{3}$

Common difference between the next two terms is also 3.

Now let's check for the next terms.

$\hookrightarrow$ $\bold{\sf{t_4}}$ - $\bold{\sf{t_3}}$

$\hookrightarrow$ $\sf{18\:-\:15}$

$\hookrightarrow$ $\sf{3}$

Common difference is constant for all the terms.

•°• The sequence 9,12,15,18 is an AP.

Calculating the 16th term :

We have the formula to calculate the n term of an AP.

Formula :

$\bold{\large{\boxed{\red{\sf{t_n\:=\:a\:+\:(n\:-\:1\:)\:d}}}}}$

Where,

• $\sf{t_n}$ = the nth term
• a = first term
• n = number of terms in an AP
• d = common difference

Block in the values ,

$\hookrightarrow$$\sf{t_{16}\:=\:9\:+\:(16\:-\:1)\:3}$

$\hookrightarrow$$\sf{t_{16}\:=\:9\:+\:15\:\times\:3}$

$\hookrightarrow$$\sf{t_{16\:=\:9\:+\:45}}$

$\hookrightarrow$$\sf{t_{16}\:=\:54}$

•°• 16th term of the AP = 54

Calculating the 24th term :

We have the same formula,

$\hookrightarrow$$\sf{t_{24}\:=\:9\:+\:(24\:-\:1)\:3}$

$\hookrightarrow$$\sf{t_{24}\:=\:9\:+\:23\:\times\:3}$

$\hookrightarrow$$\sf{t_{24}\:=\:9\:+\:69}$

$\hookrightarrow$$\sf{t_{24\:=\:78}}$

•°• 24th term of the AP = 78

Answer 2

${\huge{\sf{\underline{\underline{AnSweR:}}}}}$

•Given sequence is in AP

•$\large\sf\ a_{16}th \: term= 54$

•$\large\sf\ a_{24}th \: term= 78$

${\bold{\underline{\underline{Given:-}}}}$

•°•A sequence of AP = 9,12,15,18.....

${\bold{\underline{\underline{To \: Solve :-}}}}$

•°• We have to show that given sequence is in AP

•°• We have to also find $\sf\ a_{16}th \: and a_{24}th \: term$

_______________

We have to prove given sequence in AP ,so first we have to find 'd' of common difference and if this is regular and constant ,then the sequence is in AP

Next,

We have to find $\sf\ a_{16}th \: term \: and \: a_{24}th \: term$

for this we have to find $\sf\ a \: and \: d$

let's solve ;)

$\large\red{\boxed{\sf d = a_2 - a_1}}$

Here;

$\large\sf\star\ d = common \: difference \\ \\ \large\sf\star \ a_2 = second \: term \\ \\ \large\sf\star a_1= first \: term$

Now,

$\large\sf\star\ d_1= a_2 - a_1 =12-9 =3 \\ \\ \large\sf\star\ d_2=a_3 -a_2 = 15-12=3 \\ \\ \large\sf\star\ d_3=a_4-a_3 =18-15=3$

Hence;

$\large\sf\to\ d_1=d_2=d_3$

So, the given sequence is in AP .

Now;

$\large\red{\boxed{\sf a_n = a+(n-1)d}}$

Here;

$\large\sf\star\ a_n= n^{th} \: term$

Now solve ;

$\large\sf\star\ a= 9, d=3$

•°•$\large\sf\ Solve \: for \: a_{16}th \: term$

_______________________

$\large\sf\to\ a_{16} =</p><p>a+(n-1)d$

$\large\sf\to\ a_{16}=9+(16-1)3$

$\large\sf\to\ a_{16}=9 + 45$

$\large\bf\to\ a_{16}= 54$

•°•$\large\sf\ Solve \: for \: a_{24}th \: term$

_________________________

$\large\sf\to\ a_{24}= a+(n-1)d \\ \\ \large\sf\to\ a_{24} = 9+ (24-1)3 \\ \\ \large\sf\to\ a_{24} = 9 + 69 \\ \\ \large\bf\to\ a_ {24}= 78$