Question

Which term of the A.P $5,15,25-----$ will be $100$ more than its ${31}^{st}$ term?

• A. ${41}^{st}$
• B. ${42}^{nd}$
• C. ${40}^{th}$
• D. ${43}^{rd}$

Answer 1

The correct option is A ${41}^{st}$

The given A.P is $5,15,\mathrm{25.......}$ where the first term is $a=5$ and the common difference $d$ will be calculated by subtracting the first term from the second term as shown below:

$d=15-5=10$

Now, $31$st term$=a+30d$, substituting the values of $a$ and $d$, we get:

$31$st term$=5+(30\times 10)=5+300=305$

It is given that the A.P of the given terms is $100$ more than its $31$st term, thus, we have:

$a_n=305+100=405\Rightarrow a+(n-1)d=405\Rightarrow 5+(n-1)10=405\Rightarrow (n-1)10=405-5\Rightarrow (n-1)10=400\Rightarrow n-1=\frac{400}{10}\Rightarrow n-1=40\Rightarrow n=40+1\Rightarrow n=41$

Hence, the $41$st term of the given A.P will be $100$ more than its $31$st term.