Question

the first,second and last terms of an AP are a,b and 2a.the number of terms in AP is​
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Answer 1

Answer:

number of terms in AP = $\frac{b}{b-a}$

Step-by-step explanation:

$In \ an \ AP, \\ nth \ term \ is \ given \ by ,\\=> a(n) = a+(n-1)d \\\\ where \\ a \ is \ first \ term \\ d \ is \ common \ difference. \\\\ Given, first \ term \ ,a(1) =a \\ second \ term , a(2) =b, \\ last \ term,a(n)=2a \\\\ common \ difference,d = b-a \\\\ last \ term, a(n) = a+(n-1)d \\\\=> 2a=a+(n-1)(b-a) \\\\ => 2a-a=(n-1)(b-a) \\\\ => a=(n-1)(b-a) \\\\ =>n-1=\frac{a}{b-a} \\\\ => n =\frac{a}{b-a} +1 \\\\ =>n=\frac{a+b-a}{b-a}\\\\=>n=\frac{b}{b-a} \\\\number \ of \ terms \ in \ AP =\frac{b}{b-a}$