Math Question and Answers


Question and Answers Forum

All Questions Topic List
Mensuration Questions
Previous in All Question Next in All Question
Previous in Mensuration Next in Mensuration
Question Number 213797 by efronzo1 last updated on 17/Nov/24

	Answered by Frix last updated on 17/Nov/24
	$Σ_(n=1) ^(20) (1/(a_n a_(n+1) )) =((20)/(a_1 (a_1 +20d)))=(4/9) Σ_(n=1) ^(21) a_n =21a_1 +210d=189 ⇒ { ((a_1 =3∧d=(3/5) ⇒ a_6 =6∧a_(16) =12)),((a_1 =15∧d=−(3/5) ⇒ a_6 =12∧a_(16) =6)) :} ⇒ a_6 a_(16) =72$
	$\underset{n = 1}{\sum^{20}} \frac{1}{a_{n} a_{n + 1}} = \frac{20}{a_{1} (a_{1} + 20 d)} = \frac{4}{9}$ $\underset{n = 1}{\sum^{21}} a_{n} = 21 a_{1} + 210 d = 189$ $\Rightarrow$ ${\begin{cases} a_{1} = 3 \land d = \frac{3}{5} \Rightarrow a_{6} = 6 \land a_{16} = 12 \\ a_{1} = 15 \land d = - \frac{3}{5} \Rightarrow a_{6} = 12 \land a_{16} = 6 \end{cases}$ $\Rightarrow$ $a_{6} a_{16} = 72$
	Commented by efronzo1 last updated on 17/Nov/24

	$Why \underset{n = 1}{\sum^{20}} \frac{1}{a_{n} . a_{n + 1}} = \frac{20}{a_{1} . a_{21}} = ?$
	Commented by mehdee7396 last updated on 17/Nov/24

	$\frac{1}{a_{1} a_{2}} = \frac{1}{d} (\frac{1}{a_{1}} - \frac{1}{a_{2}})$ $\frac{1}{a_{2} a_{3}} = \frac{1}{d} (\frac{1}{a_{2}} - \frac{1}{a_{3}}), . . . ., \frac{1}{a_{20} a_{21}} = \frac{1}{d} (\frac{1}{a_{20}} - \frac{1}{a_{21}})$ $\Rightarrow \underset{1}{\sum^{20}} \frac{1}{a_{n} a_{n + 1}} = \frac{1}{d} (\frac{1}{a_{1}} - \frac{1}{a_{21}}) = \frac{20 d}{{d a}_{1} a_{21}} = \frac{20}{a_{1} a_{21}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum