ML Aggarwal Solution Class 10 Chapter 9 Arithmetic and Geometric Progressions Test
Test
Question 1
Write the first four terms of the A.P. when its first term is – 5 and the common difference is – 3.
Sol :
First 4 term of A.P. whose first term (a) = -5
and common difference (d) = -3
= -5, -8, -11, -14
Question 2
Verify that each of the following lists of numbers is an A.P., and the write its next three terms :
Sol :
(i) $0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots$
Here $a=0, d=\frac{1}{4}$
$\therefore$ Next three terms will be $1, \frac{5}{4}, \frac{3}{2}$
(ii) $5, \frac{14}{3}, \frac{13}{3}, 4, \ldots$
Here, $a=5, d=\frac{14}{3}-5=\frac{14-15}{3}=\frac{-1}{3}$
$\therefore$ Next three terms will be
$a_{2}=4-\frac{1}{3}=\frac{11}{3}$
$a_{3}=\frac{11}{3}-\frac{1}{3}=\frac{10}{3}$
$a_{4}=\frac{10}{3}-\frac{1}{3}=\frac{9}{3}=3$
i.e. $\frac{11}{3}, \frac{10}{3}, 3$
Question 3
The nth term of an A.P. is 6n + 2. Find the common difference.
Sol :
Tn of an A.P. = 6n + 2 .
$T_1$ = 6 x 1 + 2 = 6 + 2 = 8
$T_2$ = 6 x 2 + 2 = 12 + 2 = 14
$T_3 $= 6 x 3 + 2 = 18 + 2 = 20
d = 14 – 8 = 6
Question 4
Show that the list of numbers 9, 12, 15, 18, … form an A.P. Find its 16th term and the nth.
Sol :
9, 12, 15, 18, …
Here, a = 9, d = 12 – 9 = 3
or 15 – 12 = 3
or 18 – 15 = 3
Yes, it form an A.P.
$=9+15 \times 3=9+45=54$
and $\mathrm{T}_{n}=a+(n-1) d=9+(n-1) \times 3$
$=9+3 n-3=3 n+6$
Question 5
Find the 6th term from the end of the A.P. 17, 14, 11, …, – 40.
Sol :
6th term from the end of
A.P. = 17, 14, 11, …… 40
Here, a = 17, d = -3, l = -40
l = a + (n – 1 )d
$\frac{-57}{-3}=n-1$
19=n-1
n=19+1=20
$\therefore$ 6th term from the end
=l-(n-1) d
=-40-(6-1)(-3)
=-40+15=-25
Question 6
If the 8th term of an A.P. is 31 and the 15th term is 16 more than its 11th term, then find the A.P.
Sol :
In an A.P.
Let a be the first term and d be a common difference, then
Similarly,
$a_{15}=a+14 d=a+10 d+16$
14d-10d=16
$\Rightarrow 4 d=16$
$\Rightarrow d=\frac{16}{4}=4$
From
(i) $a+7 \times 4=31$
$\Rightarrow a+28=31 \Rightarrow a=31-28=3$
$\therefore a=3, d=4$
Now, A.P. will be $3,7,11,15, \ldots$
Question 7
The 17th term of anA.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, then find the wth term.
Sol :
In an A.P.
$a_{17}=2 \times a_{8}+5$
$a_{11}=43,$ find $a_{n}$
Let a be the first term and d be the common
Similarly,
$a_{17}=2 \times a_{8}+5$
a+16d=2(a+7d)+5
a+16d=2a+14d+5
-5+16a-14d=2a-a
$\Rightarrow a=2 d-5$..(ii)
From (i) and (ii)
2d-5+10d=43
$\Rightarrow 12 d=43+5=48$
$d=\frac{48}{12}=4$
But a+10d=43
$\therefore a+10 \times 4=43 \Rightarrow a+40=43$
$\Rightarrow a=43-40=3$
$\therefore a=3, d=4$
Now, $a_{n}=a+(n-1) d$
=3+4(n-1)
=3+4n-4
=4n-1
Question 8
The 19th term of an A.P. is equal to three times its 6th term. If its 9th term is 19, find the A.P.
Sol :
In an A.P.
Let a be the first term and d be the common
difference, then
a+8d=19..(i)
Similarly,
$a_{19}=3 \times a_{6}$
$\Rightarrow a+18 d=3(a+5 d)$
$a+18 d=3 a+15 d $
$\Rightarrow 3 a-a=18 d-15 d$
$\Rightarrow 2 a=3 d$..(ii)
$a=\frac{3}{2} d$
From (i)
$\frac{3}{2} d+8 d=19 $
$\Rightarrow \frac{19}{2} d=19$
$\Rightarrow d=\frac{19 \times 2}{19}=2$
and $a=\frac{3}{2} d=\frac{3}{2} \times 2=3$
$\therefore a=3, d=2$ and A.P. is $3,5,7,9, \ldots$
Question 9
If the 3rd and the 9th terms of an A.P. are 4 and – 8 respectively, then which term of this A.P. is zero?
Sol :
In an A.P.
$\Rightarrow a+2 d=4$
Similarly, a+8d=-8
Subtracting, we get
6d=-12
$ \Rightarrow d=\frac{-12}{6}=-2$
and a+2d=4
$\Rightarrow a+2 \times(-2)=4$
$\Rightarrow a-4=4$
$ \Rightarrow a=4+4=8$
Let n th term be zero, then
a+(n-1) d=0
$ \Rightarrow 8+(n-1) \times(-2)=0$
$\Rightarrow-2 n+2=-8$
$ \Rightarrow-2 n=-8-2=-10$
$\Rightarrow n=\frac{-10}{-2}=5$
$\therefore 0$ will be the fifth term.
Question 10
Which term of the list of numbers 5, 2, – 1, – 4, … is – 55?
Sol :
A.P. is 5, 2, -1, – 4, …
Which term of A.P. is -55
Let it be nth term
Here, a = 5, d = 2 – 5 = -3
$\Rightarrow-55=5+(n-1) \times(-3)$
$-55-5=-3(n-1) $
$\Rightarrow \frac{-60}{-3}=n-1$
$\Rightarrow n-1=20$
$ \Rightarrow n=20+1=21$
$\therefore-55$ is the 21 st term.
Question 11
The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is four times its 15th term.
Sol :
In an A.P.
24th term = 2 x 10th term
To show that 72nd term = 4 x 15th term
Let a be the first term and d be a common difference, then
and 72 nd term =a+71d
and 15 th term =a+14d
Substitute the value of (i), we get
a+71 d=5 d+71 d=76 d
and a+14 d=5 d+14 d=19 d
$\therefore 76 d=4 \times 19 d$
Hence 72 nd term is 4 times the 15 th term.
Question 12
Which term of the list of numbers $20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots$ is the first negative term?
$20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots$
Here, $a=20, d=19 \frac{1}{4}-20=\frac{-3}{4}$
Let n th term be first negative term
$\therefore a_{n}=a+(n-1) d$
Let n th term be first negative term, then
$a_{n}=20+(n-1)\left(\frac{-3}{4}\right)$
$\Rightarrow a_{n}=20+(n-1)\left(\frac{-3}{4}\right)$
$\Rightarrow a_{n}=20-\frac{3}{4} n+\frac{3}{4}$
Now, $a_{n}<0$ is the first negative term
$\Rightarrow 20+\frac{3}{4}-\frac{3}{4} n<0 $
$\Rightarrow \frac{83}{4}-\frac{3}{4} n<0$
$\Rightarrow \frac{83}{4}<\frac{3}{4} n \Rightarrow 83<3 n$
$\Rightarrow \frac{83}{3}<n \Rightarrow 28<n$
$\therefore 28$ th is the first negative term.
Question 13
If the pth term of an A.P. is q and the qth term is p, show that its nth term is (p + q – n)
Sol :
In an A.P.
pth term = q
qth term = p
Show that (p + q – n) is nth term
Let a be the first term and d be the common
$\therefore p$ th term =a+(p-1)d=q..(i)
and q th term =a+(q-1)d=p..(ii)
Subtracting, we get
q-p=(p-1-q+1)d=(p-q)d
$d=\frac{q-p}{p-q}=\frac{-(p-q)}{(p-q)}=-1$
From (i), $a+(p-1) \times(-1)=q$
a-p+1=q
a=q+p-1
L.H.S
nth term
=a+(n-1) d=(p+q-1)+(n-1)(-1)
=p+q-1-n+1=p+q-n=R.H.S
Question 14
How many three digit numbers are divisible by 9?
Sol :
3-digit numbers which are divisible by 9 are 108, 117, 126, 135, …, 999
Here, a = 108, d = 9 and l = 999
$\Rightarrow 999=108+(n-1) 9$
$\Rightarrow 999-108=9(n-1)$
$\Rightarrow 891=9(n-1) $
$\Rightarrow \frac{891}{9}=n-1$
$\Rightarrow n-1=99 $
$\Rightarrow n=99+1=100$
$\therefore$ There are 100 numbers or terms.
Question 15
The sum of three numbers in A.P. is – 3 and the product is 8. Find the numbers.
Sol :
Sum of three numbers of an A.P. = -3
and their product = 8
Let the numbers be
a – d, a, a + d, then
a – d + a + a + d = -3
and (a-d) a(a+d)=8
Question 16
The angles of a quadrilateral are in A.P. If the greatest angle is double of the smallest angle, find all the four angles.
Sol :
Angles of a quadrilateral are in A.P.
Greatest angle is double of the smallest
Let the smallest angle of the quadrilateral is
a+3d…..(i)
Where a-3d is the smallest
$\therefore a+3 d=2(a-3 d)$
$\Rightarrow a+3 d=2 a-6 d$
2a-a=3d+6d
$\Rightarrow a=9 d$..(ii)
But sum of angles of a quadrilateral =360°
$\therefore a-3 d+a-d+a+d+a+3 d=360^{\circ}$
$4 a=360^{\circ} \Rightarrow a=\frac{360^{\circ}}{4}=90^{\circ}$
$\therefore 9 d=a=90^{\circ} \Rightarrow d=\frac{90^{\circ}}{9}=10^{\circ}$ [From (ii)]
Greatest angle $=a+3 d=90^{\circ}+30^{\circ}=120^{\circ}$
Other angles are $=a+d=90^{\circ}+10^{\circ}=100^{\circ}$
$a-d=90^{\circ}-10^{\circ}=80^{\circ}$
and $a-3 d=90^{\circ}-30^{\circ}=60^{\circ}$
Hence angles are $60^{\circ}, 80^{\circ}, 100^{\circ}, 120^{\circ}$
Question 17
The nth term of an A.P. cannot be n² + n + 1. Justify your answer.
Sol :
nth term of an A.P. can’t be n² + n + 1
Giving some different values to n such as 1, 2, 3, 4, …
we find then
We see that,
$d=a_{2}-a_{1}=7-3=4$
$d=a_{3}-a_{2}=13-7=6$
$d=a_{4}-a_{3}=21-13=8$
We see that d is not constant
$\therefore$ It is not an A.P.
Hence, $a_{n} \neq n^{2}+n+1$
Question 18
Find the sum of first 20 terms of an A.P. whose nth term is 15 – 4n.
Sol :
Giving some different values such as 1 to 20
We get,
and so on,
$a_{20}=15-4 \times 20=15-80=-65$
Now, A.P. is $11,7,3,-1, \ldots,-65$
Here, a=11, d=-4 and n=20
$\mathrm{S}_{20}=\frac{n}{2}[2 a+(n-1) d]$
$=\frac{20}{2}[2 \times 11+(20-1)(-4)]$
=10[22-76]
=10(-54)=-540
Question 19
Find the sum :
$\mathrm{S}_{28}=\frac{28}{2}\left[2 \times 18+(28-1)\left(\frac{-5}{2}\right)\right]$
$\mathrm{S}_{28}=14\left[36+\left(27 \times \frac{-5}{2}\right)\right]$
$=14\left[36-\frac{135}{2}\right]$
$\mathrm{S}_{28}=14\left(\frac{72-135}{2}\right)=14 \times\left(\frac{-63}{2}\right)=-441$
Question 20
(i) How many terms of the A.P. – 6,$-\frac{11}{2}$ – 5,… make the sum – 25?
A.P. $=-6,-\frac{11}{2}-5, \ldots$
Here, $a=-6, d=\frac{-11}{2}+6=\frac{1}{2}$
Sum=-25
Let n term be added to get the sum -25
$\therefore \mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$-25=\frac{n}{2}\left[2 \times(-6)+(n-1)\left(\frac{1}{2}\right)\right]$
$-25 \times 2=n\left[-12+\frac{1}{2} n-\frac{1}{2}\right]$
$-50=n\left[\frac{-25}{2}+\frac{1}{2} n\right]$
$\frac{1}{2} n^{2}-\frac{25}{2} n+50=0$
$\Rightarrow n^{2}-25 n+100=0$
$\left\{\begin{array}{r}\because 100=-20 \times-5 \\ -25=-20-5\end{array}\right\}$
$\Rightarrow n^{2}-5 n-20 n+100=0$
$\Rightarrow n(n-5)-20(n-5)=0$
$\Rightarrow(n-5)(n-20)=0$
Either n-5=0, then n=5
or n-20=0, then n=20
$\therefore$ Number of terms are 5 or 20
(ii) Solve the equation 2+5+8+...+x=155
Here, a=2, d=5-2=3, l=x
Sum =155
l=a+(n-1)d
x=2+(n-1)3=2+3n-3
$\Rightarrow x=3 n-1$ ...(i)
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow 155=\frac{n}{2}[2 \times 2+(n-1) \times 3]$
$\Rightarrow 155 \times 2=n[4+3 n-3]$
$\Rightarrow 310=n(3 n+1)=3 n^{2}+n$
$\therefore 3 n^{2}+n-310=0$
$3 n^{2}-30 n+31 n-310=0$
3n(n-10)+31(n-10)=0
(n-10)(3 n+31)=0
Either n-10=0, then n=10
or 3n+31=0, then 3n=-31
$\Rightarrow n=\frac{-31}{3}$
which is not possible being negative
$\therefore n=10$
Now, x=3n-1$=3 \times 10-1=30-1=29$
[From (i)]
Question 21
If the third term of an A.P. is 5 and the ratio of its 6th term to the 10th term is 7 : 13, then find the sum of first 20 terms of this A.P.
Let a be the first term and d be the common difference
$\Rightarrow a+2 d=5$..(i)
Similarly
$a_{6}$=a+5d and $a_{10}$=a+9d
$\therefore \frac{a+5 d}{a+9 d}=\frac{7}{13}$
$ \Rightarrow 7 a+63 d=13 a+65 d$
$\Rightarrow 13 a+65 d-7 a-63 d=0$
$\Rightarrow 6 a+2 d=0 \Rightarrow 3 a+d=0$
$\Rightarrow d=-3 a$...(i)
From (i) and (ii),
a+2d=5
$\Rightarrow a+2(-3 a)=5 $
$\Rightarrow a-6 a=5$
$\Rightarrow-5 a=5 $
$\Rightarrow a=\frac{5}{-5}=-1$
and d=-3a$=-3 \times(-1)=3$
Now sum of first 20 terms
$=\frac{n}{2}[2 a+(n-1) d]$
$=\frac{20}{2}[2 \times(-1)+(20-1) \times 3]$
=10[-2+57]
$=10 \times 55=550$
Question 22
In an A.P., the first term is 2 and the last term is 29. If the sum of the terms is 155, then find the common difference of the A.P.
$l=a_{n}=a+(n-1) d$
$\Rightarrow 29=2+(n-1) d \Rightarrow 29-2=d(n-1)$
$\Rightarrow d(n-1)=27$..(i)
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$155=\frac{n}{2}[2 \times 2+27]=\frac{n}{2}[4+27]$
$155=\frac{31}{2} n $
$\Rightarrow n=\frac{155 \times 2}{31}=10$
d(n-1)=27
$\Rightarrow d(10-1)=27$
$d \times 9=27 \Rightarrow d=\frac{27}{9}=3$
Question 23
The sum of first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term.
Sol :
Sum of first 14 terms = 1505
First term (a) = 10
Find 25th term
$1505=\frac{14}{2}[2 \times 10+(14-1) d]$
$1505=7[20+13 d] \Rightarrow 20+13 d=\frac{1505}{7}$
$13 d=-20+215=195$
$d=\frac{195}{13}=15$
Now, $a_{25}=a+(n-1) d$
=10+(25-1)(15)=10+24(15)
=10+360=370
Question 24
Find the number of terms of the A.P. – 12, – 9, – 6, …, 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.
Sol :
A.P. -12, -9, -6,…, 21
If 1 is added to each term, find the sum of there terms
Last term (l)=21
$\therefore l=a_{n}=a+(n-1) d$
$\Rightarrow 21=-12+(n-1) \times 3 $
$\Rightarrow 21+12=3(n-1)$
$\Rightarrow \frac{33}{3}=n-1 \Rightarrow 11=n-1$
n=11+1=12
Now, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$=\frac{12}{2}[2 \times(-12)+(12-1) \times 3]$
$=6[-24+33]=6 \times 9=54$
By adding 1 to each term of the given A.P. the now sum will be $54+1 \times 12$
=54+12=66
Question 25
The sum of first n term of an A.P. is 3n² + 4n. Find the 25th term of this A.P.
Sol :
Sn = 3n² + 4n
Sn – 1 = 3(n – 1)² + 4(n – 1)
Now, $a_{n}=\mathrm{S}_{n}-\mathrm{S}_{n-1}$
$=\left(3 n^{2}+4 n\right)-\left(3 n^{2}-2 n-1\right)$
$=3 n^{2}+4 n-3 n^{2}+2 n+1$
=6n+1
$a_{25}=6(25)+1=150+1=151$
Question 26
In an A.P., the sum of first 10 terms is – 150 and the sum of next 10 terms is – 550. Find the A.P.
Sol :
In an A.P.
Sum of first 10 terms = -150
Sum of next 10 terms = -550, A.P. = ?
Sum of first 10 terms = -150
$\mathrm{S}_{10}=\frac{n}{2}[2 a+(n-1) d]$
$-150=\frac{10}{2}[2 a+9 d]=5(2 a+9 d)$
$\Rightarrow 10 a+45 d=-150$..(i)
$S_{20}=S_{10}+S_{10}=-150-550=-700$
$=\frac{20}{2}[2 a+19 d]$
=10(2 a+19 d)
$\Rightarrow 20a+190d=-700$...(ii)
20a+90d=-300 [Multiplying (i) by 2]
Subtracting 100d=-400
$d=\frac{-400}{100}=-4$
From (i)
$10 a+45 \times(-4)=-150$
10a-180=-150
10a=-150+180=30
$a=\frac{30}{10}=3$
A.P. is 3,-1,-5,-9, ...
Question 27
The sum of first m terms of an A.P. is 4m² – m. If its nth term is 107, find the value of n. Also find the 21 st term of this A.P.
Sol :
Sm = 4m² – m
Sn = 4n² – n
and $S_{n-1}=4(n-1)^{2}-(n-1)$
$=4\left[n^{2}-2 n+1\right]-n+1$
$=4 n^{2}-8 n+4-n+1=4 n^{2}-9 n+5$
Now, $a_{n}=\mathrm{S}_{n}-\mathrm{S}_{n-1}$
$=4 n^{2}-n-4 n^{2}+9 n-5$
=8a-5
Now, $a_{n}=107$
$\therefore 8_{n}-5=107 $
$\Rightarrow 8 n=107+5=112$
$n=\frac{112}{8}=14$
and $a_{n}=8_{n}-5$
$a_{21}=8 \times 21-5=168-5=163$
Question 28
If the sum of first p, q and r terms of an A.P. are a, b and c respectively, prove that
Sol :
Let the first term of A.P. be A and common difference be d.
Sum of the first p terms is
Sum of the first q terms is
$S_{q}=\frac{q}{2}[2 A+(q-1) d]=b$..(ii)
Sum of the first r terms is
$S_{r}=\frac{r}{2}[2 A+(r-1) d]=c$..(iii)
Now, Multiply
(i) with $\frac{q-r}{p},$(ii) with $\frac{r-p}{q},$ (iii) with
$\frac{p-q}{r}$
$\therefore \frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)$
$=\frac{(q-r)}{2}[2 \mathrm{~A}+(p-1) d]+\frac{(r-p)}{2}[2 \mathrm{~A}+(q-1) d+\frac{(p-q)}{2}[2 \mathrm{~A}+(r-1) d]$
$=\frac{2 \mathrm{~A}}{2}(q-r+r-p+p-q)+\frac{d}{2}[(q-r)(p-1)+(r-p)(q-1)+(p-q)(r-1)]$
$=0+\frac{d}{2}[(q-r) p+(r-p) q+(p-q) r-q+r-r+p-p+q]$
$=\frac{d}{2}[p q-p r+r q-p q+p r-q r+0]$
$=\frac{d}{2} \times 0=0$
Which is the required result.
Question 29
A sum of Rs 700 is to be used to give 7 cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.
What is the importance of an academic prise in students life? (Value Based)
Sol :
Total sum = Rs 700
Cash prizes to 7 students = 7 prize
Each prize is Rs 20 less than its preceding prize
d = -20, d = -20, n = 7
Now, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$700=\frac{7}{2}[2 a+(7-1)(-20)]$
$\frac{700 \times 2}{7}=2 a-120 \Rightarrow 200=2 a-120$
$2 a=200+120=320 \Rightarrow a=\frac{320}{2}=160$
$\therefore$ Cash prizes will be ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, ₹ 40
Question 30
Find the geometric progression whose 4th term is 54 and 7th term is 1458.
Sol :
In a G.P.
Similarly, $a r^{6}=1458$
Dividing, we get
$\frac{a r^{6}}{a r^{3}}=\frac{1458}{54} \Rightarrow r^{3}=27=(3)^{3}$
$\therefore r=3$
and $a r^{3}=54 \Rightarrow a \times 27=54$
$\Rightarrow a=\frac{54}{27}=2$
$\therefore a=2, r=3$
and G.P. is 2,6,18,54, ...
Question 31
The fourth term of a G.P. is the square of its second term and the first term is – 3. Find its 7th term.
Sol :
In G.P.
$a_{4}=\left(a_{2}\right)^{2}, a_{1}=-3$
$\therefore a_{n}=a r^{n-1}$
$a_{4}=a r^{3}$
$a_{2}=a r$
$a r^{3}=(a r)^{2} \Rightarrow a r^{3}=a^{2} r^{2}$
$\Rightarrow r=a \Rightarrow a_{1}=-3$
$\therefore d=-3$
$\therefore a_{7}=a r^{7}-1=a r^{6}=-3 \times(-3)^{6}$
$=-3 \times 729=-2187$
Question 32
If the 4th, 10th and 16th terms of a G.P. are x, y and z respectively, prove that x, y and z are in G.P.
In a G.P.
Show that x, y, z are in G.P.
Let a be the first term and r be the common ratio, then
Similarly, $a_{10}=a r^{9}=y$
$a_{16}=a r^{15}=z$
x, y, z are in G.P.
if $y^{2}=x y$
$y^{2}=\left(a r^{9}\right)^{2}=a^{2} r^{18}$
$x z=a r^{3} \times a r^{15}=a^{2} r^{3+15}=a^{2} r^{18}$
$\therefore y^{2}=x y$
$\therefore x, y, z$ are in G.P.
Question 33
The original cost of a machine is Rs 10000. If the annual depreciation is 10%, after how many years will it be valued at Rs 6561 ?
Sol :
Original cost of machine = Rs 10000
Since, machine depreciates at the rate of 10%
on reducing the balance,
Value of machine after one year
Value of the machine after two years
$=10,000\left(\frac{90}{100}\right) \times \frac{90}{100}$
$=10,000 \times\left(\frac{90}{100}\right)^{2}$
Thus, $10,000\left(\frac{90}{100}\right), \cdot 10,000\left(\frac{90}{100}\right)^{2} \ldots$ will
form a G.P. series with $a=10,000\left(\frac{90}{100}\right)$
and $r=\frac{9}{10}$
Let value of the machine after n years be ₹ 656
$\therefore \mathrm{T}_{n}=a r^{n-1}$
$\Rightarrow 6561=10,000\left(\frac{9}{10}\right) \times\left(\frac{9}{10}\right)^{n-1}$
$\Rightarrow \frac{6561}{10000}=\left(\frac{9}{10}\right)^{n}$
$\Rightarrow\left(\frac{9}{10}\right)^{4}=\left(\frac{9}{10}\right)^{n}$
$\Rightarrow n=4$
$\therefore n=4$
Hence, the effective life of the machine is 4 years.
Question 34
How many terms of the G.P. $3, \frac{3}{2}, \frac{3}{4}$ ,are needed to give the sum $\frac{3069}{512}$
Sol :
G.P.3, $\frac{3}{2}, \frac{3}{4}$
$S_{n}=\frac{3069}{512}$
Here, $a=3, r=\frac{1}{2}$
Let n be the number of terms
$\mathrm{S}_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$
$\frac{3069}{512}=\frac{3\left[1-\left(\frac{1}{2}\right)^{n}\right]}{1-\frac{1}{2}}$
$=\frac{3\left[1-\left(\frac{1}{2}\right)^{n}\right]}{\frac{1}{2}}$
$\frac{3069}{512}=2 \times 3\left[1-\left(\frac{1}{2}\right)^{n}\right]$
$1-\left(\frac{1}{2}\right)^{n}=\frac{3069}{512 \times 6}=\frac{1023}{1024}$
$\left(\frac{1}{2}\right)^{n}=\frac{1024-1023}{1024}=\frac{1}{1024}$
$\begin{array}{l|l}2 & 1024 \\\hline 2 & 512 \\\hline 2 & 256 \\\hline 2 & 128 \\\hline 2 & 64 \\\hline 2 & 32 \\\hline 2 & 16 \\\hline 2 & 8 \\\hline 2 & 4 \\\hline 2 & 2 \\\hline & 1\end{array}$
$\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{10}$
Comparing, we get
n=10
Question 35
Find the sum of first n terms of the series : 3 + 33 + 333 + …
Sol :
Series is
3 + 33 + 333 + … n terms
= 3[1 + 11 + 111 +…n terms]
$=\frac{3}{9}[(10-1)+(100-1)+(1000-1)+\ldots$n terms]
$=\frac{3}{9}[10+100+1000+\ldots n$ terms $-n \times 1]$
$=\frac{3}{9}\left[\left(\frac{a\left(r^{n}-1\right.}{r-1}\right)-n\right]$
$=\frac{3}{9}\left[\frac{10\left(10^{n}-1\right)}{10-1}-n\right]$
$=\frac{3}{9}\left[\frac{10}{9}\left(10^{n}-1\right)-9 n\right]$
$=\frac{3}{81}\left[10 \times 10^{n}-10-9 n\right]$
$=\frac{1}{27}\left[10^{n+1}-9 n-10\right]$
Question 36
Find the sum of the series 7 + 7.7 + 7.77 + 7.777 + … to 50 terms.
Sol :
The given sequence is 7, 7.7, 7.77, 7.777,…
= 7 + 7.7 + 7.77 + … 50 terms
= 7(1 + 1.1 + 1. 11 + … 50 terms)
$=\frac{7}{9}(10-1)+1(10-0.1)+(10-0.01)+50$ terms)
$=\frac{7}{9}[(10+10+10+\ldots 50$ terms $)-(1+0.1+0.01+\ldots 50$ terms )]
$=\frac{7}{9}[500-$ Sum of G.P. of 50 terms with a=1, r=0.1]
$=\frac{7}{9}\left[500-\frac{1\left[1-(0.1)^{50}\right]}{1-0.1}\right]$
$=\frac{7}{9}\left[500-\frac{10}{9}\left(1-\frac{1}{10^{50}}\right)\right]$
$=\frac{7}{81}\left(4500-10+10^{-49}\right)$
$=\frac{7}{81}\left[4490+10^{-49}\right]$
Question 37
The inventor of chessboard was a very clever man. He asked the king, h reward of one grain of wheat for the first square, 2 grains for the second, 4 grains for the third, and so on, doubling the number of the grains for each subsequent square. How many grains would have to be given?
Sol :
In a chessboard, there are 8 x 8 = 64 squares
If a man put 1 grain in first square,
2 grains in second square,
4 grains in third square
and goes on upto the last square, i.e. 64th square
Therefore, 1 + 2 + 4 + 8 + 16 + … 64 terms
Here, a = 1, r = 2 and n = 64
Comments
Post a Comment