Miscellaneous

Chapter 8 Class 11 Sequences and Series
Serial order wise

### Transcript

Misc 11 Find the sum of the following series upto n terms: (ii) .6 +.66 +. 666 +β¦ .6 +.66 +. 666 +β¦ to n terms Here, 0.66/0.6 = (66/100)/(6/10) = 1.1 & 0.666/0.66 = (666/1000)/(66/100) = 1.009 Thus, (ππππππ π‘πππ)/(πΉπππ π‘ π‘πππ) β  (πβπππ π‘πππ)/(ππππππ π‘πππ) i.e. common ratio is not same β΄ This is not a GP We need to find sum Sum = 0.6 + 0.66 + 0.666 + β¦ to n terms = 6 [0.1 + 0.11 + 0.111 +β¦ to n terms ] = 6 [0.1 + 0.11 + 0.111 +β¦ to n terms ] Multiplying & dividing by 9 = 6/9 Γ 9[0.1 + 0.11 + 0.111 +β¦ to n terms ] = 2/3 [0.9 + 0.99 + 0.999 +β¦ to n terms ] = 2/3 [ (9/10) + (99/100)+ (999/1000)+β¦ to n terms ] = 2/3 [((10 β 1)/10) + ((100 β 1)/100)+ ((1000 β 1)/1000)+ β¦to n terms ] = 2/3[(1 β 1/10)+ (1 β 1/100)+ (1 β 1/1000) + β¦ to n terms ] = 2/3 [(1 + 1 +β¦β¦.to n terms) β (1/10 " + " 1/100 " + " 1/1000 " + β¦ to n terms" )] = 2/3 [n Γ 1 β (1/10 " + " 1/100 " + " 1/1000 " + β¦ to n terms" )] Now, a = 1/10, r = 1/10 For, r < 1 i.e. Sn = (a(1 βπ^π))/(1 β π) Putting value of a = 1/10 & r = 1/10 = (1/10 (1 β(1/10)^π ))/(1 β 1/10) = (1/10 (1 β(1/10)^π ))/(9/10) = 1/10[1 β (10)^(βπ)] Γ 10/9 = 1/9[1 β (10)^(βπ)] Thus, (1/10 " + " 1/100 " + " 1/1000 " + β¦ to n terms" ) = 1/9[1 β (10)^(βπ)] Now, Sum = 2/3 [n β (1/10 " + " 1/102 " + " 1/103 " + β¦ to n terms" )] Substitute (1/10 " + " 1/100 " + " 1/1000 " + β¦ to n terms" ) = 1/9[1 β (10)^(βπ)] in (1) = 2/3 [n β 1/9[1 β (10)^(βπ)] = 2/3n β 2/3 Γ1/9[1 β (10)^(βπ)] = 2/3 n β 2/27[1 β (10)^(βπ)] Hence , 0.6 + 0.66 + 0.666 + β¦ to n terms = 2/3 n β 2/27[1 β (10)^(βπ)]