Hello students welcome back so in this module will be discussing arithmetic geometric progression let us see what does arithmetic geometric progression means and shot it is called as AGP now students let us say i am given an AP and its terms are a1 a2 a3 till an And AGP term are b1 ,b2,b3 till bn then we can say the sequence which is obtained by multiplying the corresponding terms of an AP and AGP that is the fall term of AP into the first term of GP that is A1 b1second term of a AP second term of GP multiplied that is a2 B2 and then so on till AN BN would be called as arithmetic geometric Progression lets us from the AP is given as a ,a plus d a plus 2d and so on and GP is given that is B, BR& Br square till Br power of n minus 1 then multiplying the corresponding terms of these two sequence will give me a sequence which would look like this, this will become an AGP .students you can see over here and nTh term of AGP can be written as the multiplication of in nth term of AP and the end , term GP this one is the most important concept of here is to find the sum of N term in Agp let’s say I denote some of term in a AGP Sn then sn would be this some of these terms over here this students very important step of over here is I’m going to multiply complete series by the common ratio of GP as we used to do it in terms of in GP topic that is geometric progression this from multiplying complete series by R that is a common ratio so i have multiplied by R and shifted The terms like the this students no don’t worry i have written it in this fashion now i’m going to subtract these two series so when i’m going to subtract these two series this turns note over here the terms start cancelling out ABR cancels from the second and in the first expression similarly the second term third term and so on till interval Observe like this so what i’m going to subtract his students 1 – R into Sn would be written like this you can see over his students I obtained this expression and now i’m taking DB common from the second term and I obtained over you can see db into R plus R squared till R to the power n minus 1 and you can see this is AGP with first term R and common ratio also are so using the sum of N times in GP that is a into r to the power of n minus 1 upon r minus 1 we obtain this result and are now i’m dividing by 1 minus R to sum of n terms is open like this the students we don’t have to remember this formula we have to remember this method let us move further and find some of in finite terms in a AGP now students of common ratio better – 1to1 number of terms are in finite we have already found the sum of n terms in AGP and now it’s already given that R is between – 1to 1 number of times n is infinity therefore students R to the power n-1 when we were in understand this concept previously also that R to the power n -1towards zero and R to the power also towards zero using these two results in the above expression I’ll obtain our to the power n minus 1 and r to the power of n from the last two terms cancels out gives me 0 and hence some of in finite terms you can see obtained as ab upon 1 minus r plus dbr upon 1 minus r square So students let us use these concepts in solving some question lets his students have to find the sum of these terms let’s check is this an AGP . look at the first term 1 into 2, 2nd 2*2 Square, 3 *2cube observe the terms of each and every term that is one two three and so on that is AP and the second terms that is 2,2 square, 2 cubed till to the power of hundred that is AGP there for students these terms are written in AGP and to find some of these term students we already understood we multiply complete series by the Common ratio of GP over here the common ratio of GPS students too some 1 to multiply the complete sequence by two so student multiply the complete sequence by 2 shift read the terms i’m going to subtract the two series students so i’m subtracting s minus 2’s becomes minus s and now the rest of the terms will be one into the first term as it is from second term onwards that’s where 2 Square, 2 cube, 2 power 4, two power of hundred minus the last term is written As it is now students you can see from first term to the hundredth term it’s AGP hence applying the sum of a hundred times in GP that is a into R to the power of n minus 1 upon r minus 1 that is two minus one minus the last term as it is that is hundred to two power of 101 its now simple calculations students doing the calculation i obtain this final result I hope his students you understood this concept thank you
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