Question

The 11^th term and the 21^st term of an A.P. are 16 and 29 respectively, then find the 41^th term of that A.P.

Answer 1

It is given that,
a₁₁ = 16 and a₂₁ = 29
We know that,
an=a+(n-1)d

a11=a+(11-1)d

⇒16=a+10d

⇒a=16-10d ...1

a21=a+(21-1)d

⇒29=a+20d

⇒29=16-10d+20d from 1

⇒29-16=10d

⇒d=$\frac{13}{10}$ ...2

Putting the value of d in 1, we get

a=16-10x1.3=3

∴ a41=a+(41-1)d

=3+40x1.3

=55
Hence, the 41^th term of the A.P. is 55.