Question

In a harmonic progression $t_{1},t_{2},t_{3},\ldots \ldots \ldots ..,$ it is given that $t_{5}=20$ and $t_{6}=50.$ If $S_{n}$ denotes the sum of first $n$ terms of this, then the value of $n$ for which $S_{n}$ is maximum is

• A. $6$
• B. $7$
• C. $9$
• D. $10$

Answer 1

Answer: (A)

Let the first term be $a$ and the common difference be $d$ , then
$20=\frac{1}{a + 4 d}$ and $50=\frac{1}{a + 5 d}$
On solving, we get,
$d=-\frac{3}{100}$ and $a=\frac{17}{100}$
Hence, $t_{r}=\frac{100}{20 - 3 r}>0\Rightarrow r < \frac{20}{3}\Rightarrow r_{m a x}=6$
So, sum is maximum for $n=6,$ as after this terms are negative