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
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
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!!
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.
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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)