Insert two numbers between 1 7 and 1 13 so that the resulting sequence is a H.P. Solution:

Let the required numbers be 1 H_1 and 1 H_2 . 1 7 , 1 H_1 , 1 H_2 , 1 13 are in H.P. 7, H_1, H_2 and 13 are in A.P. t_1=a=7 and t_4=a+3 ~d=13 7+3 d=13 3 ~d=6 d=2 H_1=t_2=a+d=7+2=9 and H_2=t_3=a+2 ~d=7+2(2)=11 1 9 and 1 11 are the required numbers to be inserted between 1 7 and 1 13 so that the resulting sequence is a H.P.

Insert two numbers between 1 7 and 1 13 so that the resulting seq…

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Insert two numbers between $\frac{1}{7}$ and $\frac{1}{13}$ so that the resulting sequence is a H.P. Solution:

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Answer

Let the required numbers be $\frac{1}{\mathrm{H}_1}$ and $\frac{1}{\mathrm{H}_2}$.
$\therefore \frac{1}{7}, \frac{1}{\mathrm{H}_1}, \frac{1}{\mathrm{H}_2}, \frac{1}{13}$ are in H.P.
$\therefore 7, \mathrm{H}_1, \mathrm{H}_2$ and 13 are in A.P.
$\therefore \mathrm{t}_1=\mathrm{a}=7$ and $\mathrm{t}_4=\mathrm{a}+3 \mathrm{~d}=13$
$\therefore 7+3 d=13$
$\therefore 3 \mathrm{~d}=6$
$\therefore \mathrm{d}=2$
$\therefore \mathrm{H}_1=\mathrm{t}_2=\mathrm{a}+\mathrm{d}=7+2=9$
and $\mathrm{H}_2=\mathrm{t}_3=\mathrm{a}+2 \mathrm{~d}=7+2(2)=11$
$\therefore \frac{1}{9}$ and $\frac{1}{11}$ are the required numbers to be inserted between $\frac{1}{7}$ and $\frac{1}{13}$ so that the resulting sequence is a H.P.

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Y	500	800	870	599	625	650	905	700	850	824
X	82	105	99	90	108	124	90	90	111	89
Y	970	609	990	735	600	735	729	840	999	780
X	112	100	87	92	91	82	96	120	121	122
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Sequences and Series (p-1) — Maths (commerce) STD 11 Commerce / Arts — Question

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X	110	88	91	115	97	85	85	91	120	95
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Y	970	609	990	735	600	735	729	840	999	780
X	112	100	87	92	91	82	96	120	121	122
Y	638	850	630	720	695	923	555	810	805	526