Q.1. If the sum of reciprocals of first 11 terms of an HP series is 110, find the 6^th term of HP.

Explanation: Now from the above HP formulae, it is clear the reciprocals of first 11 terms will make an AP. The sum of first 11 terms of an AP = [2a + (11 – 1) d] 11/2 = 110
⇒ 2a + 10d = 20 ⇒ a + 5d = 10
Now there are 2 variables, but a + 5d = T₆ in an AP series. And reciprocal of 6^th term of AP series will give the 6^th term of corresponding HP series. So, the 6^th term of HP series is 1/10

Q.2. Find the 4th and 8th term of the series 6, 4, 3, …

Explanation: Consider 1/6, /14, 1/3, ...
Here T₂ – T₁ = T₃ – T₂ = 1/12
Therefore 1/6, 1/4, 1/3 is in A.P.
4th term of this Arithmetic Progression = 1/6 + 3 × 1/12 = 1/6 + 1/4 = 5/12,
Eighth term = 1/6 + 7 × 1/12 = 9/12.
Hence the 8th term of the H.P. = 12/9 = 4/3 and the 4th term = 12/5.

Q.3. The 2nd term of an HP is 40/9 and the 5th term is 20/3. Find the maximum possible number of terms in H.P.

Explanation : If a, a + d, a + 2d, a + 3d, ……. are in A.P. then [image] are in H.P.
Now [image] and [image]
Solving these two equations we get [image] and [image]
Now [image]
Hence the maximum terms that this H.P. can take is 10.

Q.4. If the first two terms of a harmonic progression are 1/16 and 1/13, find the maximum partial sum?

Answer: Option A

Explanation : The terms of the HP are [image]
So the maximum partial sum is [image]

Q.5. The second term of an H.P. is [image] and the fifth term is [image] . Find the sum of its 6^th and the 7^th term.

Explanation: If a, a + d, a + 2d, a + 3d, ……. are in A.P. then [image] are in H.P.
Now [image]
and [image]
Solving these two equations, we get [image] and [image]
Now [image]
And [image]
So the sum of the 6^th and the 7^th term of H.P. is [image]

Q.6. If the sixth term of an H.P. is 10 and the 11th term is 18 Find the 16th term.

Answer: Option C