Question

The 11th11th term and the 21st21st term of an A.P. are $16$ and $29$ respectively, then find $n$ such that $t_n=55.$

• A. $40$
• B. $41$
• C. $42$
• D. $43$

Answer 1

The correct option is B $41$

Let $a$ and $d$ be the first term and common difference of given AP respectively.
Given $t_{11}=16and{t}_{21}=29$
$\Rightarrow a+10d=16$ ...(1)
$a+20d=29$ ...(2)
Subtraction (1) from (2), we get
$10d=13$
$d=\frac{13}{10}$
Putting $d=\frac{13}{10}$ in (1), we get

$a+10\times \frac{13}{10}=16\Rightarrow a=3$

Now, $t_n=55\Rightarrow a+(n-1)d=55\Rightarrow 3+(n-1)\times \frac{13}{10}=55$
$\Rightarrow (n-1)\times \frac{13}{10}=52$
$\Rightarrow n-1=52\times \frac{10}{13}$
$n=41$