Sequence and Series – Test 1

The required series will be 105,110,115………….995

therefore,last term = 995

a+(n-1)d=995

105+(n-1)5=995

(n-1)5=890

n-1=890/5

n=178+1=179

sums of the series = n/2[a+l]

=179/2[105+995]

=98450

a=-4 and l=146

let the number of terms be n

sum=7171

n/2[a+l]=7171

n/2[-4+146]=7171

142 n=14342

n=101

let the first terms of G.P be a, then its second term=a-2

common ratio i.e. r=a-2/a

sum of infinity=50

a/1-r=50

a/1-(a-2)/a=50

a/a-a+2/a=50

a=10

r=10-2/10=8/10=4/5

therefore,the required series is 10,8,32/5……..

let the 5 parts be a-2d , a-d , a , a+d , a+2d

then , (a-2d)+(a-d)+(a)+(a+d)+(a+2d)=30

5a=30

a=6

further,a-2d/a+2d=2/3

6-2d/6+2d=2/3

18-6d=12+4d

6=10d

d=3/5

substituting the values of a and d ,the 5 parts are:

6-2×3/5 , 6-3/5 , 6 , 6+3/5 , 6+2×3/5

=24/5 , 27/5 , 6 , 33/5 , 36/5

$$\begin{gathered} let a^{1/x} = b^{1/y} = c^{1/z} = k \\ then,a=k^x , b=k^y , c=k^z \\ \sin ce a,b,c are in G.P \\ then b^2=ac \\ (k^y{)}^2=k^x{k}^z \\ {k}^{2y}=k^{x+z} \\ 2y=x+z \\ y=x+z/2 \\ this showa that x,y,z are in A.P \end{gathered}$$

The required series will be 252,255,258………….999

therefore,last term = 999

a+(n-1)d=999

252+(n-1)3=999

(n-1)3=747

n-1=747/3

n=249+1=250

sums of the series = n/2[a+l]

=250/2[252+999]

=156375

$$\begin{gathered} given,S_n=3n^2–n \\ S1=3x1–1=3–1=2 \\ S1=t1 \\ t1=2 \end{gathered}$$

$$\begin{gathered} \sin ce the series is an AP with a=2,d=5 \\ l=297 \\ a+(n–1)d=297 \\ 2+(n–1)5=297 \\ n=59+1 \\ n=60 \\ sum=n/2[a+l] \\ =60/2[2+297] \\ =8970 \end{gathered}$$

since the ball was dropped from a height of 100 m ,so it travelled 100 m

then,it went up to 100×4/5=80 m and dropped.so it travelled 2×80 m

then this way the total distance travelled would be:

$$\begin{gathered} here,a=200 , d=25 and S_n=9450 \\ Assume that the contract time is over run for n days.then \\ {S}_n=\frac{n}{2}[2a+(n–1\left)d\right] \\ 9450=\frac{n}{2}[2x200+(n–1\left)25\right] \\ 18900=n[400+25n–25] \\ 18900=n(375+25n) \\ 18900=375n+25n^2 \\ 25n^2+375n–18900=0 \\ {n}^2+15n–756=0 \\ {n}^2+36n–21n–756=0 \\ n(n+36)–21(n+36)=0 \\ (n–21)(n+36)=0 \\ n=21 or n=–36 \\ hence,no.of days cant be negative \\ so n=21 days \end{gathered}$$

$$\begin{gathered} T_1=a \\ {T}_2=a+(2–1)d=a+d \\ {T}_7=a+(7–1)d=a+6d \\ \sin ce T_1,T_2 and T_7 are in GP \\ a+d/a=a+6d/a+d \\ a+2/a=a+12/a+2 \\ (a+2{)}^2=a(a+12) \\ {a}^2+4a+4=a^2+12a \\ 12a–4a=4 \\ 8a=4 \\ a=1/2 \\ therfore,second term \frac{1}{2}+2=5/2 \end{gathered}$$

$$\begin{gathered} given,a=3000x12=36000 \\ n=20 \\ d=1000x12=12000 \\ {S}_n=\frac{n}{2}[2a+(n–1\left)d\right] \\ {S}_{20}=\frac{20}{2}[2x36000+(20–1\left)12000\right] \\ =10x300000 \\ =Rs 3000000 \end{gathered}$$

$$\begin{gathered} let A_1,A_2,A_3,A_4 are the 4 A.M‘s 3 and 18. \\ then the numbers 3,A_1,A_2,A_3,A_4 18 from an AP. \\ whose 1st term is 3 and 6th term is 18 \\ {T}_n=a+(n–1)d \\ 18=3+(6–1)d \\ 5d=15 \\ d=3 \\ then,A_1=3+3=6 \\ A_2=6+3=9 \\ A_3=9+3=12 \\ A_4=12+3=15 \end{gathered}$$

$$\begin{gathered} x = 1+\frac{1}{3}+\frac{1}{3^2}+................\infty \\ x = a/1–r = 1/1–\frac{1}{3} \\ x = 3/2 \\ y = 1+\frac{1}{4}+\frac{1}{4^2}+..............\infty \\ y = 1/1–\frac{1}{4}= 4/3 \\ xy =\frac{ 3}{2}x\frac{4}{3}=2 \end{gathered}$$

$$\begin{gathered} The investment made by a person is forming an AP with d=100 and S_{10}=54500. \\ if a be the 1st term, we have \\ {S}_n=\frac{n}{2} [2a+(n–1\left)d\right] \\ 54500=10/2[2a+(10–1\left)100\right] \\ 54500/5=2a+900 \\ 2a=10900–900 \\ a=5000=value of certificates purchased in the 1st year \end{gathered}$$

since (x+1), 3x, (4x+2) are in AP

2(3x)=(x+1)+(4x+2)

since a,b,c are in AP, 2b=a+c

6x=5x+3

x=3

$$\begin{gathered} given AM=10 and GM=8 \\ \frac{a+b}{2}=10 and \sqrt{ab=8} \\ a+b=20.......i \\ ab=64........ii \\ b=20–a from i \\ sub b in ii \\ a(20–a)=64 \\ 20a–a^2–64=0 \\ {a}^2–20a+64=0 \\ {a}^2–16a–4a+64=0 \\ a(a–16)–4(a–16)=0 \\ (a–16)(a–4)=0 \\ now, a=16, b=4 or a=4, b=16 \\ numbers are [16,4] \end{gathered}$$