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Chapter 5,, e.x 5:3 question number 3

Answer» Solution :- (i) given that a=5 , d=3, an=50 we know nth term of an AP, an= a+(n-1)d =50=5+(n-1)3 =50-5=(n-1)3 =45=(n-1)3 =45/3=n-1 =15=n-1 =n=15+1=n=16Now,sum of n terms, Sn=n/2 [a+an] Sn=16/2 [5+50] sn=16/2 ×55 Sn=8×55=440 . (ii) given that,a=7,a13 =35 we know nth term of an AP, an =a+n-1 =a13=7+(13-1)d=35=7=12d=35-7=12d=28=12d=d=28/12=7/3Now,sum of n terms Sn =n/2 [a+an] S13=13/2 [7+25] S13=13/2×42S13=273.(iv) given that a3=15, S10 =125 we know, nth term of an AP, An=a+(n-1)= a3=a+(3-1)d--------------> (1)Sum of n terms Sn=n/2 [2a+(n-1)d]=S10=10/2 [2a+(10-1)d]=S10=5(2a+9d)-------------> (2)multiplying equ.(1) by (2) we get 30=2a+4d ----------> (3)Subtracting equ.(2) from equ (3) we get 2a+4d=30 2a+9d=25 (-) (-) _____________ -5d =5 d=___5__= -1 -5subtracting the value of d in the equ (1)we get 15=a+2(-1) =15=a+2=a=15+2=a=a=17Therefore a10 =a+(10-1)da10=17+(10-1)(-1)a10=17+(10-1)(-1)a10=17+9×(-1)a10=17-9a10=8 thus , d= -1 and a10=8 (v) given that, d=5, Sq=75 we know sum of n terms, Sn=n/2 [2a+(n-1)d] = Sq=9/2 [2a+(9-1)]=75=9/2 [2a+40] =75×2=9[2a+40]=150×18a+360=18a=150-360=18a= -210=a= -210/18=-35/3we know, an =a+(9-1)da9=-35/3 +(9-1)5a9= -35/3 +40a9= -35+120/3 =85/3 thus a=-35/3 and a9=85/3(vi) given a=2, d=8, Sn=90we know,sum of n terms,Sn =n/2 [2a+(n-1)] =90=n/2 [2 (2)+(n-1)4]=90= n/2×2 [2+(n-1)4]=90=n [2+(4n-4]=90= n[4n-2]=90=4n^ - 2n=4n^-2n-90=0=2n^-n-45=0=2n^-10n+9n-45 (splitting the middle term)=2n(n-5)+9(n-5)=0=(n-5)(Zn+9)=0=n-5=0 we know nth term of an AP = an =an=2+(5-1)8=an=2+2 4×8=an= 2+32=an=34 thus, n=5 and an=34(viii) given that an=4,d=2, Sn=-14we know nth term of an AP,an =an=a+(n-1)2=4=a+2n-2=6=a+2n=a=6-2n ------------>(1)we know sum of n terms Sn=n/2[2a+(n-1)]= -14=n/2×2[a+(n-1)]= -14=n[a+(n-1)] -------------->subtracting equ (1) in equ.(2) we get -14= [6-2n+n-1]= -14=n [5-n]= -14=5n-n2=n2-5n-14=0=n2-7n +2n-14=0=n(n-7)+2 (n-7)=0=(n-7)(n+2)=0=n-7=0=n=7 subtracting the value of\'n\' in the equ (1)we get =a=6-2(7)=a=6-14=a= -8 thus n=7 and a= -8(ix) given that a=3, n=8 and s=192 we know the sum of n terms of an AP Sn= n/2[a+an]=192=8/2 [3+an]=192=8/2[3+an]=192/4=[3+an]=192/4=3+an=48=3+an=an=48-3=an=45we know nth term of an APan=a+(n-1)d=45=3+(8-1)d=45=7d=d=6 thus d=6


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