Class 11

Math

Algebra

Sequences and Series

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

- $40$
- $41$
- $42$
- $43$

Given $t_{11}=16andt_{21}=29$

$⇒a+10d=16$ ...(1)

$a+20d=29$ ...(2)

Subtraction (1) from (2), we get

$10d=13$

$d=1013 $

Putting $d=1013 $ in (1), we get

$a+10×1013 =16⇒a=3$

Now, $t_{n}=55⇒a+(n−1)d=55⇒3+(n−1)×1013 =55$

$⇒(n−1)×1013 =52$

$⇒n−1=52×1310 $

$n=41$

$⇒(n−1)×1013 =52$

$⇒n−1=52×1310 $

$n=41$