On substituting n = 1, 2, 3, 4, 5, we get first 5 terms
\(a_1=1. \dfrac{1^2+5}{4}\)
\( a_1=\dfrac{6}{4}=\dfrac{3}{2}\)
\( a_2=2. \dfrac{2^2+5}{4}\)
\( a_2=\dfrac{18}{4}=\dfrac{9}{2}\)
\( a_3=3. \dfrac{3^2+5}{4}\)
\( a_3= \dfrac{42}{4}=\dfrac{21}{2}\)
\( a_4=4. \dfrac{4^2+5}{4}\)
\( a_4=\dfrac{84}{4}=21\)
\( a_5=5. \dfrac{5^2+5}{4}\)
\( a_5= \dfrac{150}{4}=\dfrac{75}{2}\)
Hence, the required terms are \( \dfrac{3}{2}\), \( \dfrac{9}{2}\),\( \dfrac{21}{2}\) , 21 and \( \dfrac{75}{2}\).
Answered by Abhisek | 1 year agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.
Find the geometric means of the following pairs of numbers:
(i) 2 and 8
(ii) a3b and ab3
(iii) –8 and –2
Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).