In a harmonic progression t1,t2,t3, ldots ldots ldots .., it is given that t5=20 and t6=50. If Sn denotes the sum of first n terms of this, then the value of n for which Sn is maximum is

Question

In a harmonic progression t1,t2,t3, ldots ldots ldots .., it is given that t5=20 and t6=50. If Sn denotes the sum of first n terms of this, then the value of n for which Sn is maximum is (A) 6 (B) 7 (C) 9 (D) 10

Tardigrade Admin · Accepted Answer

Let the first term be a and the common difference be d , then 20=(1/a + 4 d) and 50=(1/a + 5 d) On solving, we get, d=-(3/100) and a=(17/100) Hence, tr=(100/20 - 3 r)>0⇒ r < (20/3)⇒ rm a x=6 So, sum is maximum for n=6, as after this terms are negative

Q. In a harmonic progression $t_{1}, t_{2}, t_{3}, \dots\dots\dots ..,$ 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

$6$

$7$

$9$

$10$

Q. In a harmonic progression t1​,t2​,t3​,……….., it is given that t5​=20 and t6​=50. If Sn​ denotes the sum of first n terms of this, then the value of n for which Sn​ is maximum is

6

7

9

10

Let the first term be a and the common difference be d , then 20=a+4d1​ and 50=a+5d1​ On solving, we get, d=−1003​ and a=10017​ Hence, tr​=20−3r100​>0⇒r<320​⇒rmax​=6 So, sum is maximum for n=6, as after this terms are negative

Q. In a harmonic progression $t_{1}, t_{2}, t_{3}, \dots\dots\dots ..,$ 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

$6$

$7$

$9$

$10$