Question

1 , 3 , 5 , 7 ,...... , 99


Find the no. of terms in AP.​

Answer 1

Given :-

$\sf{ a\:=\: 1}$

$\sf{d \:=\: 2}$

$\sf{\ a_{n} \:=\:99 }$

To find :-

$\sf{Number\:of\: terms(n)}$

Formula used !

★${\boxed{\sf{\ a_{n}\:=\:a+(n-1)d}}}$★

Solution :-

${\sf{\ a_{n}\:=\:a+(n-1)d}}$

${\sf{99\:=\:1+(n-1)2}}$

${\sf{99-1\:=\:(n-1)2}}$

${\sf{\cancel{98}\:=\:(n-1)\cancel{2}}}$

${\sf{n-1\:=\:49}}$

${\sf{n\:=\:50}}$

$\bf{So\:no.\:of\:terms\:in\:this\:Ap\:is\:50 }$

Answer 2

$\huge\bf{Given}$

$\tt{1,3,5,7,9.........,99}$

$\huge\bf{Simplifying \: it }$

$\bf{Step\:by\:step \: solution :-}$

$\bf{In \: This\: AP }$

$\bf{A\: = \: 1 }$

$\bf{D\: = 3-1\:=\:2 }$

$\bf{A_{n} \: = \: 99}$

$\bf{As\: We\: Know }$

$\bf{\boxed{ A_{n}\:=\:a\:+\:(n-1)\:d}}$

$\bf{ 99\:=\:1\:+\:(n-1)\:2}$

$\bf{\frac{99-1}{2}=\:n-1}$

$\bf{\frac{98}{2}=\:n-1}$

$\bf{\frac{\cancel98}{\cancel2}=\:n-1}$

$\bf{49}=\:n-1$

$\bf{49+1}=\:n$

$\bf{50}=\:n$

$\huge\bf{Hope\:It\:Help\:You}$