 Hello and welcome to the session. In this session we discussed the following question which says show that 3 to the power 1 upon 3 into 27 to the power 1 upon 9 into 233 to the power 1 upon 27 into 2187 to the power 1 upon 81 into and so on up to infinity is equal to 3. Before we move on to the solution let's see what is Arithmetico geometric series. The standard form of the series is a plus a plus d the whole into r plus the whole into r square plus and so on plus a plus m minus 1 whole into d and this whole into minus 1 plus the plan B corresponding term from AP and the GP a plus a plus d plus a plus 2d plus a plus m minus 1 into d r to the power m minus 1 that we used for this question. Let's proceed with the solution now. We are supposed to prove upon 3 into 27 to the power 1 upon 43 to the power 1 upon 27 into 2187 to the power 1 upon 81 into and so on up to infinity is equal to 3. First of all we consider let 3 to the power 1 upon 3 into 27 to the power 1 upon 9 into 243 to the power 1 upon 27 into 2187 to the power 1 upon 81 to take log on both sides so log x is equal to log of 1 3 into 27 to the power 1 upon 9 into 3 to the power 1 upon 27 into 1 upon 81 into and so on of the algorithm which says log x 1 x 2 to the base a is equal to log x 1 to the base a plus log x 2 to the base a so this log can be applied to this y hand x is equal to log 3 to the power 1 upon 3 plus log 27 to the power 1 upon 9 plus log 3 to the power 1 upon 27 plus to the power 1 upon 81 when this after 27 we get 3 to the power 3 yes we have log x is equal to log 3 to the power 1 upon 3 plus log 3 to the power 3 upon 9 3 is equal to 3 to the power 5 log 3 to the power 5 upon 287 is equal to 3 to the power of 7 so we can write log 3 to the power of 7 upon 81 plus the next log algorithm that we can use here is log of x 1 to the power n this to the base a is equal to m into log x 1 to the base a so using this log on the right hand side for each term we would get log x is equal to 1 upon 3 into log 3 log 3 into 7 into log 3 upon 81 log 3 further we have x is equal to log 3 common and this log 3 into 7 upon 81 plus equal to this upon 3 into 7 plus 7 upon 81 matrix series where each term is found by multiplying the corresponding terms of the 80 where the 1 plus 3 plus 5 where the GP is 1 upon 3 plus 1 upon 9 1 upon 27 plus 1 upon 81 upon 3 is the common ratio of this GP say we multiply this s by 1 upon 3 so we have 1 upon 3 s is equal to into 1 upon 3 minus 1 upon 9 plus 3 upon 9 27 1 upon 3 s we get 1 minus 1 upon 3 this whole into s is equal to 1 upon 3 minus 0 is 1 upon 3 plus 3 upon 9 minus 1 upon 9 upon 27 is 2 upon 27 plus minus 5 upon 81 is 2 upon 81 so that means we now have s is equal to 1 upon 3 plus 37 upon 3 which is less than 1 is equal to a upon which in this case is 1 upon 3 equal to 1 upon 3 and the geometric series which would be 2 upon 9 which is the first term upon 1 minus r that is 1 minus 1 upon 3 that is we now further have 2 upon 3 s is equal to 1 upon 3 plus and 3 3 times is 9 is equal to plus 1 upon 3 which is equal to 2 upon 3 upon 3 cancels so this means we have s is equal to 1 that is we get log x is equal to 1 into log 3 is equal to log 3 and this means that x is equal to 3 and we have assumed x to be equal to therefore we can say upon 3 7 to the power 43 to the power 1 upon 27 into 2 1 8 7 to the power 1 upon 81 of what we were supposed to this question.