Gujarati
8. Sequences and Series
easy

यदि $\frac{1}{3}$ और $\frac{1}{{24}}$ के मध्य दो समान्तर माध्य पद ${A_1}$ व ${A_2}$ हों, तब ${A_1}$ व ${A_2}$ का मान होगा

A

$\frac{7}{{72}},\,\frac{5}{{36}}$

B

$\frac{{17}}{{72}},\,\frac{5}{{36}}$

C

$\frac{7}{{36}},\,\frac{5}{{72}}$

D

$\frac{5}{{72}},\,\frac{{17}}{{72}}$

Solution

(b) यहाँ  $\frac{1}{3},\;{A_1},\;{A_2},\;\frac{1}{{24}}$ समान्तर श्रेणी में होंगे, 

 $\therefore $ ${A_1} – \frac{1}{3} = \frac{1}{{24}} – {A_2}$

$ \Rightarrow $ ${A_1} + {A_2} = \frac{3}{8}$……$(i)$

अब चूँकि ${A_1}$, $\frac{1}{3}$व ${A_2}$ का समान्तर माध्य है।

अत: $2{A_1} = \frac{1}{3} + {A_2} $

$\Rightarrow 2{A_1} – {A_2} = \frac{1}{3}$……$(ii)$

$(i)$ व $(ii)$ से, ${A_1} = \frac{{17}}{{72}}$ और ${A_2} = \frac{5}{{36}}$.

वैकल्पिक : सूत्र ${A_m} = a + \frac{{m(b – a)}}{{n + 1}}$

जहाँ $n = 2,\;a = \frac{1}{3},\;b = \frac{1}{{24}}$

$\therefore $ ${A_1} = \frac{1}{3} + \frac{{ – 7/24}}{3} = \frac{{17}}{{72}}$

 ${A_2} = \frac{1}{3} + \frac{{ – 14/24}}{3} = \frac{{10}}{{72}} = \frac{5}{{36}}$.

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.