12)its sumation of n2 n varin frm 11 -20
=(n(n+1)(2n+1))/6 here n=10.......so itz divisible by 5
Ques1) The lengths of three unequal rectangular block arein G.P. The volume of the block is 216 cm3 and the total surface area is 252 cm2 . Show that the length of the longest edge is 12 cm.
Ques2)Show that 11 2 + 12 2 +13 2 + .............+ 20 2 is an odd integer divisible by 5.
Ques3) The consecutive odd integers whose sum is 45 2 - 21 2 are
(a) 43, 45, ...... 75
(b) 43,45,.........79
(c) 43, 45,.........85
(d) 43, 45,.........89.
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9 Answers
621 is easy.. i wonder why you are finding it difficult
take the sides as a/r, a, ar
you will get a=6 and r=2
622 use the fact that the sum of these numbers is 1^2+2^2.....20^2 - (1^1+2^2.... 10^2)
or that the sum of squares of 5 consecutive integers squares is always divisible by 5
62question three is a great one..
what is teh sum of first 45 odd numbers?
1
1Q.2) just use :( n(n+1)(2n+1)/6)1-20 - [n(n+1)(2n+1)/6]1-10
get the product n get the answer.
1Q3) the sum of first 45 odd nos. = 452
and the sum of first 21 odd nos. = 212
so their difference is the sum of the 22nd to 45th odd nos.
hence answer is d
341similar problem list appearing in goiit: http://www.goiit.com/posts/list/algebra-a-p-g-p-974577.htm
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