Harmonic Progression (H.P.) — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Harmonic Progression (H.P.) MCQs with step-by-step solutions (20 questions). Part of Sequences and Series. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Harmonic Progression (H.P.) · easy · theory
A sequence is a Harmonic Progression (H.P.) if:
A. the reciprocals of its terms form a G.P.
B. the reciprocals of its terms form an A.P. ✓ Correct
C. its terms form a G.P.
D. its terms form an A.P.
Solution: By definition, $a_1, a_2, \dots$ is an H.P. iff $\tfrac{1}{a_1}, \tfrac{1}{a_2}, \dots$ is an A.P.
Q2 — Harmonic Progression (H.P.) · easy · theory
The reciprocals of the H.P. $1, \tfrac13, \tfrac15, \tfrac17, \dots$ form the A.P.:
A. $1, 2, 3, 4, \dots$
B. $2, 4, 6, \dots$
C. $1, 3, 5, 7, \dots$ ✓ Correct
D. $1, \tfrac13, \tfrac15, \dots$
Solution: Reciprocals are $1, 3, 5, 7, \dots$, an A.P. with common difference $2$.
Q3 — Harmonic Progression (H.P.) · easy · numerical
The harmonic mean of $4$ and $6$ is:
A. $4.8$ ✓ Correct
B. $5.2$
C. $5$
D. $4.5$
Solution: H.M. $= \dfrac{2ab}{a+b} = \dfrac{2 \cdot 24}{10} = 4.8$.
Q4 — Harmonic Progression (H.P.) · easy · numerical
The harmonic mean of $2$ and $8$ is:
A. $3.2$ ✓ Correct
B. $5$
C. $4$
D. $3.5$
Solution: H.M. $= \dfrac{2 \cdot 16}{10} = 3.2$.
Q5 — Harmonic Progression (H.P.) · medium · theory
If $a, b, c$ are in H.P., then $b$ equals:
A. $\dfrac{2ac}{a+c}$ ✓ Correct
B. $\sqrt{ac}$
C. $\dfrac{ac}{a+c}$
D. $\dfrac{a+c}{2}$
Solution: The middle term of a three-term H.P. is the harmonic mean: $b = \dfrac{2ac}{a+c}$.
Q6 — Harmonic Progression (H.P.) · medium · numerical
The $5$-th term of the H.P. $6, 3, 2, \dots$ is:
A. $\tfrac32$
B. $1$
C. $\tfrac{5}{6}$
D. $\tfrac{6}{5}$ ✓ Correct
Solution: Reciprocals $\tfrac16, \tfrac13, \tfrac12, \dots$ form an A.P. with $d = \tfrac16$; the $n$-th reciprocal is $\tfrac{n}{6}$, so $T_5 = \tfrac{6}{5}$.
Q7 — Harmonic Progression (H.P.) · medium · numerical
If two harmonic means are inserted between $1$ and $\tfrac14$, they are:
A. $\tfrac23$ and $\tfrac12$
B. $\tfrac12$ and $\tfrac13$ ✓ Correct
C. $\tfrac13$ and $\tfrac14$
D. $\tfrac14$ and $\tfrac15$
Solution: Reciprocals form an A.P. $1, 2, 3, 4$; the inserted reciprocals $2, 3$ give H.M.s $\tfrac12, \tfrac13$.
Q8 — Harmonic Progression (H.P.) · medium · numerical
Which term of the H.P. $\tfrac15, \tfrac19, \tfrac{1}{13}, \dots$ is $\tfrac{1}{101}$?
A. $20$-th
B. $24$-th
C. $26$-th
D. $25$-th ✓ Correct
Solution: Reciprocals $5, 9, 13, \dots$ ($d = 4$): $5 + 4(n-1) = 101 \Rightarrow n = 25$.
Q9 — Harmonic Progression (H.P.) · hard · numerical
If the harmonic mean of two numbers is $4$ and their arithmetic mean is $5$, their geometric mean is:
A. $2\sqrt{5}$ ✓ Correct
B. $4.5$
C. $4$
D. $3$
Solution: Using $G^2 = A \cdot H = 5 \times 4 = 20$, we get $G = 2\sqrt{5}$.
Q10 — Harmonic Progression (H.P.) · medium · numerical
The $7$-th term of the H.P. $\tfrac12, \tfrac15, \tfrac18, \dots$ is:
A. $\tfrac{1}{20}$ ✓ Correct
B. $\tfrac{1}{17}$
C. $\tfrac{1}{23}$
D. $\tfrac{1}{19}$
Solution: Reciprocals $2, 5, 8, \dots$ ($d = 3$): $T_7 = 2 + 6(3) = 20$, so the term is $\tfrac{1}{20}$.
Q11 — Harmonic Progression (H.P.) · medium · theory
If $a, b, c$ are in H.P., then $\dfrac{a - b}{b - c}$ equals:
A. $\dfrac{b}{a}$
B. $\dfrac{a}{c}$ ✓ Correct
C. $1$
D. $\dfrac{c}{a}$
Solution: A standard H.P. property: for $a, b, c$ in H.P., $\dfrac{a-b}{b-c} = \dfrac{a}{c}$.
Q12 — Harmonic Progression (H.P.) · medium · numerical
The harmonic mean of $6$ and $12$ (the middle term making $6, b, 12$ an H.P.) is:
A. $7.5$
B. $10$
C. $8$ ✓ Correct
D. $9$
Solution: H.M. $= \dfrac{2 \cdot 6 \cdot 12}{18} = \dfrac{144}{18} = 8$.
Q13 — Harmonic Progression (H.P.) · hard · numerical
In an H.P., the $5$-th term is $\tfrac{1}{16}$ and the $8$-th term is $\tfrac{1}{25}$. Its general ($n$-th) term is:
A. $\dfrac{1}{4n}$
B. $\dfrac{1}{2n+3}$
C. $\dfrac{1}{3n-1}$
D. $\dfrac{1}{3n+1}$ ✓ Correct
Solution: Reciprocals: $a + 4d = 16, a + 7d = 25 \Rightarrow d = 3, a = 4$, so the $n$-th reciprocal is $3n + 1$.
Q14 — Harmonic Progression (H.P.) · medium · theory
For the reciprocals of an A.P. $2, 4, 6, 8, \dots$, the resulting H.P. is:
A. $1, \tfrac12, \tfrac13, \dots$
B. $2, 4, 6, 8, \dots$
C. $\tfrac{1}{2}, \tfrac{1}{3}, \tfrac14, \dots$
D. $\tfrac12, \tfrac14, \tfrac16, \tfrac18, \dots$ ✓ Correct
Solution: Taking reciprocals of an A.P. gives an H.P.: $\tfrac12, \tfrac14, \tfrac16, \tfrac18, \dots$
Q15 — Harmonic Progression (H.P.) · medium · numerical
If three harmonic means are inserted between $\tfrac12$ and $\tfrac16$, the middle one is:
A. $\tfrac15$
B. $\tfrac12$
C. $\tfrac14$ ✓ Correct
D. $\tfrac13$
Solution: Reciprocals $2, 3, 4, 5, 6$; the middle inserted reciprocal is $4$, giving H.M. $\tfrac14$.
Q16 — Harmonic Progression (H.P.) · medium · numerical
The harmonic mean of the three numbers $1, 2, 4$ is:
A. $2$
B. $\dfrac{7}{3}$
C. $\dfrac{12}{7}$ ✓ Correct
D. $\dfrac{7}{12}$
Solution: H.M. $= \dfrac{3}{1 + \tfrac12 + \tfrac14} = \dfrac{3}{7/4} = \dfrac{12}{7}$.
Q17 — Harmonic Progression (H.P.) · hard · numerical
If $a, b, c$ are in H.P. with $a = 4$ and $c = 12$, then $b$ is:
A. $9$
B. $8$
C. $6$ ✓ Correct
D. $5$
Solution: $b = \dfrac{2 \cdot 4 \cdot 12}{4 + 12} = \dfrac{96}{16} = 6$.
Q18 — Harmonic Progression (H.P.) · hard · theory
If $A$, $G$, $H$ are the A.M., G.M. and H.M. of two positive numbers, then $A$, $G$, $H$ are in:
A. G.P. (i.e. $G^2 = AH$) ✓ Correct
B. H.P.
C. no fixed relation
D. A.P.
Solution: It is a standard result that $G^2 = A \cdot H$, so $A, G, H$ are in G.P.
Q19 — Harmonic Progression (H.P.) · hard · numerical
The $10$-th term of an H.P. whose first two terms are $\tfrac13$ and $\tfrac15$ is:
A. $\tfrac{1}{19}$
B. $\tfrac{1}{23}$
C. $\tfrac{1}{17}$
D. $\tfrac{1}{21}$ ✓ Correct
Solution: Reciprocals $3, 5, \dots$ ($d = 2$): $T_{10} = 3 + 9(2) = 21$, so the term is $\tfrac{1}{21}$.
Q20 — Harmonic Progression (H.P.) · medium · theory
Among the A.M., G.M. and H.M. of two distinct positive numbers, the smallest is:
A. the G.M.
B. they are equal
C. the H.M. ✓ Correct
D. the A.M.
Solution: Since $A \ge G \ge H$ with equality only when the numbers are equal, the H.M. is the least for distinct numbers.