Question

plzz help..........​

Answer 1

Solution :-

Given Arthemetic Progressions to us is

➤ 21 , 42 , 63 , 84 , ____

We are required to find next term , and we can solve it by two methods .

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } MethoD\:1:-}}}}}$

The AP is 21 , 42 , 63 , 84 , ____

The common difference is 42 - 21 = $\red{\sf 21 }$ . So , the next will be obtained by adding common difference to the previous term .

Here that previous term is 84 . So , next term would be ;

= $\tt 84 + 21$

= $\boxed{\purple{\bf 105 }}$

__________________________________

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } MethoD\:2:-}}}}}$

Here we will use the formula to find the nth term of AP when first term common difference and n is given .

$\tt Here , the \:terms\:are$

$\sf First \:term= 21$

$\sf Common\: Difference=21$

$\sf nth \:term = 5$

$\bf \qquad\qquad \underline{\boxed{\bf The\: formula\:is}}$

$\large{\boxed{\red{\bf \green{\dag}\:T_n\:=\:a+(n-1)d}}}$

$\tt Using\: this\: formula$

$\tt :\implies T_n=a+(n-1)d$

$\tt :\implies T_5= 21+(5-1)21$

$\tt :\implies T_5 = 21+4\times21$

$\tt :\implies T_5=21+84$

$\underline{\boxed{\red{\tt{\longmapsto\:Term_5\:\:=105}}}}$

$\boxed{\green{\pink{\dag}\: \sf Hence\:the\: required\: term \:is\:105.}}$

Answer 2

Answer:

Solution :-

Given Arthemetic Progressions to us is

➤ 21 , 42 , 63 , 84 , ____

We are required to find next term , and we can solve it by two methods .

$\large{\underline{\underline{\red{\tt{\pink{\leadsto } MethoD\:1:-}}}}}$

The AP is 21 , 42 , 63 , 84 , ____

The common difference is 42 - 21 = $\orange{\sf 21 }$ . So , the next will be obtained by adding common difference to the previous term .

Here that previous term is 84 . So , next term would be ;

= $\tt 84 + 21$

= $\boxed{\blue{\bf 105 }}$

__________________________________

$\large{\underline{\underline{\purple{\tt{\pink{\leadsto } MethoD\:2:-}}}}}$

Here we will use the formula to find the nth term of AP when first term common difference and n is given .

$\tt Here , the \:terms\:are$

$\sf First \:term= 21$

$\sf Common\: Difference=21$

$\sf nth \:term = 5$

$\bf \qquad\qquad \underline{\boxed{\bf The\: formula\:is}}$

$\large{\boxed{\blue{\bf \green{\dag}\:T_n\:=\:a+(n-1)d}}}$

$\tt Using\: this\: formula$

$\tt :\implies T_n=a+(n-1)d$

$\tt :\implies T_5= 21+(5-1)21$

$\tt :\implies T_5 = 21+4\times21$

$\tt :\implies T_5=21+84$

$\underline{\boxed{\purple{\tt{\longmapsto\:Term_5\:\:=105}}}}$

$\boxed{\pink{\orange{\dag}\: \sf Hence\:the\: required\: term \:is\:105.}}$