Arithematic Progressions Upto nth term

Arithematic Progressions Upto nth term
Solved Problems
Problem1 : If \(a,b,c \in {R^ + }\) form an A.P, then prove that \(a + \frac{1}{{bc}},\,b + \frac{1}{{ac}},\,c + \frac{1}{{ab}}\) are also form an A.P   
Sol  : Given a,b,c are in A.P
    \( \Rightarrow \frac{a}{{abc}},\,\,\frac{b}{{abc}},\frac{c}{{abc}}\) are in A.P
   \(\Rightarrow \frac{1}{{bc}},\,\,\frac{1}{{ca}},\frac{1}{{ab}}\)  are in A.P
   \(\Rightarrow a + \frac{1}{{bc}},\,b + \frac{1}{{ca}},\,c + \frac{1}{{ab}}\)  will also be in A.P, since sum of two A.P's is also an A.P
Problem 2 : If 100 times the 100^th term of an A.P with non-zero common difference equals the 50 times of 50^th term, then find 150^th term of this A.P.
Sol : Given 100 T₁₀₀ = 50 T_{50
\(\Rightarrow 100(a + 99d) = 50\left( {a + 49d} \right)\)
\(\Rightarrow 2a + 198d = a + 49d\)
\( \Rightarrow a + 149d = 0\)
\(\Rightarrow {T_{150}} = 0\)
Problem 3 :The number of natural numbers lying between 100 and 1000, which are multiples of 5 is
Sol : The numbers which are multiples of 5 between 100 and 1000 are 105,110,115,.........995
This is a A.P with first term a = 105, common difference = d=5
 Now, Last term = tn = 995
\(\Rightarrow 105 + (n - 1)5 = 995\)
\(\Rightarrow (n - 1)5 = 890\)
\(\Rightarrow n - 1 = \frac{{890}}{5}\)
\(\Rightarrow n - 1 = 178\)
\(\Rightarrow n = 179\)
The number of numbers = 179}