The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.
It clearly forms a sequence in A.P.
Where, the first term, a = 1
Common difference, d = 2
Now,
a + (n -1)d = 2001
1 + (n-1)(2) = 2001
2n – 2 = 2000
2n = 2000 + 2 = 2002
n = 1001
We know,
Sn = \( \dfrac{n}{2}[2a+(n-1)d]\)
= \( \dfrac{1001}{2}[2\times 1+(1001-1)\times 2]\)
= \( \dfrac{1001}{2}[2+1000 \times 2]\)
= \( \dfrac{1001}{2}\times 2002\)
= 1002001
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 \( \dfrac{81}{2}\).