1. Class 11
2. Important Question for exams Class 11

Transcript

Example 15 Find the sum of the sequence 7, 77, 777, 7777, ... to n terms. 7, 77, 777, 7777, ... n terms Here, 77/7 = 11 & 777/77 = 10.09 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 = 7 + 77 + 777 + 7777 + ...upto n terms = 7(1 + 11 + 111 + β¦. β¦ upto n terms) = 7(1 + 11 + 111 + β¦. β¦ upto n terms) Multiplying & dividing by 9 = 7/9 [9(1 + 11 + 111 + β¦upto n term) = 7/9 [9 + 99 + 999 + 9999 + β¦upto n terms] = 7/9 [(10 β 1) + (100 β 1) + (1000 β 1) +β¦upto n terms] = 7/9 [(10 + 100 + 1000 + β¦.n terms) β 1 β 1 β 1 ββ¦upto n terms] = 7/9 [(10 + 100 + 1000 + β¦.n terms) β (1 + 1 + 1 +β¦upto n terms)] = 7/9 [(10 + 100 + 1000 + β¦.n terms) β n Γ 1] = 7/9 [(10 + 100 + 1000 + β¦.n terms) β n] Now, a = 10, r = 10 For, r > 1 i.e. Sn = (a(π^πβ 1))/(π β 1) Putting value of a = 10 & r = 10 Sn = (10(γ10γ^π β 1))/(10 β 1) Sn = (10(γ10γ^π β 1))/9 Now substituting this value in (1) Sum = 7/9 [(10 + 102 + 103 + β¦ upto n terms) β n] Sum = 7/9 [(10(γ10γ^π β 1))/9 " β n" ] Thus, 7, 77, 777, 7777, ...upto n terms = 7/9 [(10(γ10γ^π β 1))/9 " β n" ]

Class 11
Important Question for exams Class 11