1for 3rd
use sum of terms equidistant from beginning and end is equal
in this case a1 + a19 = a2 +a18
=== 112 ..ok?
now u can do this i think!!
Q1)If x,y and z are in AP,prove that 1/√x+√y ,1/√z+√x ,1/√y+√z are in AP
Q2)Sum the series to n terms:
(1/1+√x) +(1/1-x)+(1/1-√x)+.......
Q3)Find the sum of first 19 terms of the AP a1,a2,a3,.........if it is known that a1+a8+a12+a19=224
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6 Answers
1
341\frac{1}{\sqrt x + \sqrt y} + \frac{1}{\sqrt y + \sqrt z} = \frac{\sqrt y - \sqrt x}{d} + \frac{\sqrt z - \sqrt y}{d} = \frac{\sqrt z - \sqrt x}{d} = \frac{2}{\sqrt x + \sqrt z} (d is of course the common difference)
This establishes that \frac{1}{\sqrt x + \sqrt y}, \frac{1}{\sqrt x + \sqrt z}, \frac{1}{\sqrt y + \sqrt z} are in A.P.
1thrz bit mistake in #2
in this case a1 + a19 = a2 +a18
=== 112
in ques a8+a12 instead.......bt the same method is applied..........thx
62see if you have typed the 2nd problem correctly
Q3)Find the sum of first 19 terms of the AP a1,a2,a3,.........if it is known that a1+a8+a12+a19=224
observe that...
a1 + a19 = a8+a12
now sum of an AP is sum of 1st and last term divided by number of terms..
11for the second quest. the terms are in a.p. with common difference √x /(1-x) therefore the sum to nterms would be
Sn = [n{√x(n-3) + 2}]/2(1-x)