7 + 77 + 777 + … to n terms.
Let us take 7 as a common term so we get,
7 [1 + 11 + 111 + … to n terms]
Now multiply and divide by 9 we get,
\( \dfrac{7}{9}\) [9 + 99 + 999 + … n terms]
\( \dfrac{7}{9}\)[(10 – 1) + (102 – 1) + (103 – 1) + … + (10n – 1)]
\( \dfrac{7}{9}\) [(10 + 102 + 103 + … +10n)] –\( \dfrac{7}{9}\) [(1 + 1 + 1 + … to n terms)]
So the terms are in G.P
Where, a = 10, r = \( \dfrac{10^2}{10}\) = 10, n = n
By using the formula,
Sum of GP for n terms =
\( \dfrac{7}{9}\) [\( \dfrac{10}{9}\) (10n – 1) – n]
\( \dfrac{7}{81}\)[10 (10n – 1) – n]
\( \dfrac{7}{81}\)(10n+1 – 9n – 10)
Answered by Sakshi | 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 \( \dfrac{81}{2}\).