Chapter 5,, e.x 5:3 question number …

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=16 Now,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=7/3 Now,sum of n terms Sn =n/2 [a+an] S13=13/2 [7+25] S13=13/2×42 S13=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 -5 subtracting the value of d in the equ (1)we get 15=a+2(-1) =15=a+2=a =15+2=a =a=17 Therefore a10 =a+(10-1)d a10=17+(10-1)(-1) a10=17+(10-1)(-1) a10=17+9×(-1) a10=17-9 a10=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=-35/3 we know, an =a+(9-1)d a9=-35/3 +(9-1)5 a9= -35/3 +40 a9= -35+120/3 =85/3 thus a=-35/3 and a9=85/3 (vi) given a=2, d=8, Sn=90 we 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=-14 we 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=45 we know nth term of an AP an=a+(n-1)d =45=3+(8-1)d =45=7d =d=6 thus d=6