Question

If the $3rd,6th$ and last term of a H.P. are $\frac{1}{3},\frac{1}{5},\frac{3}{203},$ then the number of terms is equal to

• A. $100$
• B. $102$
• C. $99$
• D. $101$

Answer 1

The correct option is A $100$

If $\frac{1}{3},\frac{1}{5},\frac{3}{203}$ are $3rd,6th$ and last term of a H.P.

Then $3,5,\frac{203}{3}$ are $3rd,6th$ and last term of a A.P.

$n$th term of AP , $a_n=a+(n-1)d$

So, $a+2d=3$ .......... (1)

and $a+5d=5$ .......... (2)

From (1) and (2),

$3d=2$

$d=\frac{2}{3}$

From (1),

$a=3-2d$

$a=3-\frac{4}{3}=\frac{5}{3}$

Let $\frac{203}{3}$ is $n$th term of AP

then

$a+(n-1)d=\frac{203}{3}$

$\frac{5}{3}+(n-1)\frac{2}{3}=\frac{203}{3}$

$5+2(n-1)=203$

$2(n-1)=198$

$n-1=99$

$n=100$