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If the sum of reciprocals of first 11 terms of H.P series is 110, then find the ${{6}^{th}}$ term of H.P
( a ) $\dfrac{1}{10}$
( b ) $\dfrac{2}{5}$
( c ) $\dfrac{3}{10}$
( d ) $\dfrac{4}{9}$

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Last updated date: 13th Jun 2024
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Answer
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Hint: To solve this question, what we will find the sum of the first 11 terms of A.P and then put 110 equal to the formula of the sum of A.P and then from there we will get the relation between a and d. then we will find the ${{6}^{th}}$ term of A.P and then we will reciprocal the value which will be equal to ${{6}^{th}}$ term of H.P.

Complete step-by-step solution:
Before we solve this question, let’s see what are Arithmetic Progression and Harmonic Progression.
Arithmetic Progression is a series whose consecutive elements have a constant difference.
Denoted as a, a+d, a+2d, a+3d, ….., a+(n-1)d, where a = first term, d = difference between consecutive terms and n = number of terms in A.P
Harmonic progression is a series containing, reciprocal of terms of an A.P
Denoted as $\dfrac{1}{a}'\dfrac{1}{a+d}'\dfrac{1}{a+3d}'......'\dfrac{1}{a+(n-1)d}$ , $\dfrac{1}{a}$ = first term, n = number of terms in A.P
Now, it is given that the reciprocal of the first 11 terms of H.P terms sums 110. And obviously, the reciprocal of H.P will give us A.P.
So, sum of first 11 terms of an A.P = 110
Sum of first n terms $=\dfrac{n}{2}\left( 2a+(n-1)d \right)$
$110=\dfrac{n}{2}\left( 2a+(n-1)d \right)$
As, n = 11
So, $110=\dfrac{11}{2}\left( 2a+10d \right)$
On solving, we get
$10=\left( a+5d \right)$…….)( i)
Now, ${{n}^{th}}$ term of an A.P = a + (n-1)d
So, ${{6}^{th}}$ term of A.P = a + (6-1)d = a + 5d
From, ( i ), ${{6}^{th}}$ term of A.P = 10
As, terms of H.P is reciprocal of terms of A.P,
So, ${{6}^{th}}$ term of H.P = $\dfrac{1}{10}$.

Note: As we see on the question is asked in very clever language so one must know the concept of A.P and H.P series deeply. All formulas related to A.P and H.P must be known to solve the questions. Do not forget to convert terms of A.P in terms of H.P by doing reciprocal of terms of A.P.