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Question

# Find the sum of the following geometric series : (i) 0.15 + 0.015 + 0.0015 + ..... to 8 terms ; (ii) √2+1√2+12√2+..... to 8 terms ; (iii) 29−13+12−34+..... to 5 terms. (iv) (x+y)+(x2+xy+y2)+(x3+x2y+xy+y3)+.... to n terms ; (v) 35+452+353+454+..... to 2n terms ; (vi) a1+i+a(1+i)2+a(1+i)3+....+a(1+i)n (vii) 1,−a,a2,−a3,........ to n terms (a≠1) (viii) x3,x5,x7,.... to n terms (ix) √7,√21,3√7,.... to n terms

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Solution

## (i) 0.15 + 0.015 + 0.0015 + .... upto 8 terms Here, a = 0.15 and r=a2a1=0.0150.15=110. S8=a(1−r81−r) =0.15⎛⎝1−(110)81−110⎞⎠=0.15(1−1108110)=16(1−1108) (ii) √2+1√2+12√2+.... to 8 terms ; Here the first term of the series is a=√2 and the common ratio is r=1√2√2=12 Thus the sum of the G.P. up to 8th terms is : S8=a(1−r8)1−r=√2(1−(12)8)1−12 =2√2(1−1256)=255√2128 (iii) 29−13+12−34+.... to 5 terms. a=29,r=−1329=−13×92=−32,n=5 S5=(1−r5)1−r =29(1−(−32)5)1−(−32) =29(1+24332)1+32 =29(275)32×25=5572 (iv) (x+y)+(x2+xy+y3)+(x3+x2y+xy2+y3)+..... =1x−y{(x2−y2)+(x3−y3)+...to ∞}.......... [∵xn−ynx−y=xn−1+xn−2y+...+yn−1] =1x−y{(x2+x3+....to ∞)−(y2+y3+....to ∞)} =1x−y{x21−x−y21−y} =1x−y{x2−x2y−y2+xy2(1−x)(1−y)} =x+y−xy(1−x)(1−y) (v) 35+452+353+454+.... to 2n terms ; The series can be written as : 3(15+153+155+....n terms)+4(152+154+156+....n terms) For the first part a=15 and the common ratio r=152=125 Thus the sum is : 3(15+153+155+...n terms)=315(1−(125)n)1−125 =58(1−152n) For the second part a=125 and common ratio r=125 then 4(152+154+156+.....n terms)=4.125(1−(125)n)1−125=16(1−152n) Thus the sum is : 35+452+352+....2n terms =58(1−152n)+16(1−152n) =58+16(1−152n) =15+424(1−152n) =1924(1−152n) (vi) a1+i+a(1+i)2+a(1+i)3+....+a(1+i)n a=a1+i, r=a(1+i)2a1+i=11+iSn=a(1−rn)1−r=a1+i(1−(11+i)n)1−11+i=a1+i×1+i(−i)(1−(1+i)n)=−ai(1−(1+i)−n) (vii) 1,−a,a2,−a3,.... to n terms (a≠1) Rewriting the sequence and sum we get, Sum=1−a+a2−a3+a4−a5+.... Here, r = - a and first term = 1 Sum=[1−(−a)n]1+a (viii) x3,x5,x7,..... to n terms Here the first term of the G.P, is a = x3 and the common ratio is r=x5x3=x2 Thus the sum of the G.P. is x3+x5+x7+.... to n terms =x3((x2)n−1)x2−1=x2n−1x2−1 (ix) √7,√21,3√7,.... to n terms Here the first term of the G.P. is a = √7 (ix) √7,√21,3√7,.... to terms Here the first term of the G.P. is a = √7 and the common ratio is r=√21√7=√3 Thus the sum of the G.P. is : √7+√21+3√7+.... to n terms =√7((√3)n−1)√3−1=√7(3n2−1)√3−1

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