 Hello and welcome to the session. In this session, we discussed the following question that says sum up 6 plus 66 plus 666 plus and so on to n terms Before we go on to the solution, let's see what would be the sum of n terms of a GP That is a geometric progression. This is equal to a into 1 minus r to the power n and the small upon 1 minus r Where this r is not equal to 1 where this a is the first term of the GP and R is the common ratio And of course n is the number of terms of the GP This is the key idea that we use in this question. Let's now move on to the solution We are supposed to sum up the CD's to n terms. So first of all, we suppose let S be equal to 6 plus 66 plus 666 plus and so on plus tn minus 1 plus tn where this tn is the nth term of the series and tn minus 1 is the n minus 1th term Now we will solve the series by the method of difference as you can see that the successive difference of the series For a GP. So we will find the sum to n terms of the series by the method of difference So after this we have Also s is equal to 6 plus 66 plus and so on plus tn minus 2 plus tn minus 1 plus tn subtracting the 2 We get that is subtracting this and this We get s minus s which is 0 is equal to 6 minus 0 is 6 plus 66 minus 6 is 60 plus 666 minus 66 which would be 600 Plus and so on plus tn minus minus 1 Then 0 minus tn would be minus tn or you can say we have 0 is equal to 6 plus 60 plus 600 plus and so on n terms GP and This whole minus tn Now shifting this tn to the left hand side we get tn is equal to 6 plus 60 plus 600 plus and so on up to n terms Now as this is a GP and we know that the sum of n terms of a GP is given by a into 1 minus r to the power n upon 1 minus r Where this r is the common ratio a is the first term and this r should not be equal to 1 Now the common ratio for this GP would be given by 60 upon 6 Which is 10 which is same as 600 upon 60 which is 10 Now as this 10 is greater than 1 So some of n terms of the GP would be given by a into r to the power n minus 1 upon r minus 1 Since this r is greater than 1 So here we would have tn is equal to a which is 6 Into r to the power n and this r is 10 So 10 to the power n minus 1 this whole upon 10 minus 1 so we get Tn is equal to 6 upon 9 this whole into 10 to the power n minus 1 Now 3 2 times is 6 and 3 3 times is 9 So we have tn is equal to 2 upon 3 and this whole into 10 to the power n minus 1 so does we now have the nth term of the series Now Tn minus 1 is given by 2 upon 3 this whole into 10 to the power n minus 1 minus 1 Then in the same way Tn minus 2 is equal to 2 upon 3 into 10 to the power n minus 2 minus 1 The whole that of in place of n we put here n minus 2 similarly So on and we have t2 that is the second term is Equal to 2 upon 3 into 10 to the power 2 minus 1 And t1 is equal to 2 upon 3 into 10 to the power 1 minus 1 the whole Now next we would add these terms to get the sum of the series to n terms Which we have denoted by s So now we have s would be equal to t1 Plus t2 plus and so on plus tn minus 2 plus tn minus 1 plus tn So this would be equal to 2 upon 3 into 10 minus 1 plus 2 upon 3 into 10 square minus 1 the whole plus and so on plus 2 upon 3 into 10 to the power n minus 2 minus 1 the whole plus 2 upon 3 into 10 to the power n minus 1 minus 1 the whole plus 2 upon 3 into 10 to the power n minus 1 the whole Now we can take 2 upon 3 common here. So 2 upon 3 into 10 plus 10 square plus and so on plus 10 to the power n minus 2 plus 10 to the power n minus 1 plus 10 to the power n minus n. This n is because of the 1 being subtracted n times. This is the sum of n terms of the given series. Now again as you can see that this would be a GP. So we find out the sum of this GP and right here. So this is equal to 2 upon 3 into now sum of this GP is a which is 10 into x to the power n and r in this case is 10. So 10 raised to the power n minus 1 and this whole upon 10 minus 1 which is 9 and this minus n. Now have s is equal to 2 upon 3 into 10 upon 9 into 10 to the power n minus 1 the whole minus n. The sum of the n terms of the given series. So sum of the n terms given series is equal to 2 upon 3 into 10 upon 9 into 10 raised to the power n minus 1 the whole minus n the whole. So this is our final answer. This completes the session. Hope you have understood the solution of this question.