Sequences and Series
NCERT Class 11 Chapter 9 – Sequences and Series
Exercise 9.1 Page No: 180
Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:
1. an = n (n + 2)
Solution:
Given,
nth term of a sequence an = n (n + 2)
On substituting n = 1, 2, 3, 4, and 5, we get the first five terms
a1= 1(1 + 2) = 3
a2= 2(2 + 2) = 8
a3= 3(3 + 2) = 15
a4= 4(4 + 2) = 24
a5= 5(5 + 2) = 35
Hence, the required terms are 3, 8, 15, 24, and 35.
2. an = n/n+1
Solution:
Given nth term, an = n/n+1
On substituting n = 1, 2, 3, 4, 5, we get
Hence, the required terms are 1/2, 2/3, 3/4, 4/5 and 5/6.
3. an = 2n
Solution:
Given nth term, an = 2n
On substituting n = 1, 2, 3, 4, 5, we get
a1 = 21 = 2
a2 = 22 = 4
a3 = 23 = 8
a4 = 24 = 16
a5 = 25 = 32
Hence, the required terms are 2, 4, 8, 16, and 32.
4. an = (2n – 3)/6
Solution:
Given nth term, an = (2n – 3)/6
On substituting n = 1, 2, 3, 4, 5, we get
Hence, the required terms are -1/6, 1/6, 1/2, 5/6 and 7/6..
5. an = (-1)n-1 5n+1
Solution:
Given nth term, an = (-1)n-1 5n+1
On substituting n = 1, 2, 3, 4, 5, we get
Hence, the required terms are 25, –125, 625, –3125, and 15625.
6.
Solution:
On substituting n = 1, 2, 3, 4, 5, we get first 5 terms
Hence, the required terms are 3/2, 9/2, 21/2, 21 and 75/2.
Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:
7. an = 4n – 3; a17, a24
Solution:
Given,
nth term of the sequence is an = 4n – 3
On substituting n = 17, we get
a17 = 4(17) – 3 = 68 – 3 = 65
Next, on substituting n = 24, we get
a24 = 4(24) – 3 = 96 – 3 = 93
8. an = n2/2n ; a7
Solution:
Given,
nth term of the sequence is an = n2/2n
Now, on substituting n = 7, we get
a7 = 72/27 = 49/ 128
9. an = (-1)n-1 n3; a9
Solution:
Given,
nth term of the sequence is an = (-1)n-1 n3
On substituting n = 9, we get
a9 = (-1)9-1 (9)3 = 1 x 729 = 729
10.
Solution:
On substituting n = 20, we get
Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:
11. a1 = 3, an = 3an-1 + 2 for all n > 1
Solution:
Given, an = 3an-1 + 2 and a1 = 3
Then,
a2 = 3a1 + 2 = 3(3) + 2 = 11
a3 = 3a2 + 2 = 3(11) + 2 = 35
a4 = 3a3 + 2 = 3(35) + 2 = 107
a5 = 3a4 + 2 = 3(107) + 2 = 323
Thus, the first 5 terms of the sequence are 3, 11, 35, 107 and 323.
Hence, the corresponding series is
3 + 11 + 35 + 107 + 323 …….
12. a1 = -1, an = an-1/n, n ≥ 2
Solution:
Given,
an = an-1/n and a1 = -1
Then,
a2 = a1/2 = -1/2
a3 = a2/3 = -1/6
a4 = a3/4 = -1/24
a5 = a4/5 = -1/120
Thus, the first 5 terms of the sequence are -1, -1/2, -1/6, -1/24 and -1/120.
Hence, the corresponding series is
-1 + (-1/2) + (-1/6) + (-1/24) + (-1/120) + …….
13. a1 = a2 = 2, an = an-1 – 1, n > 2
Solution:
Given,
a1 = a2, an = an-1 – 1
Then,
a3 = a2 – 1 = 2 – 1 = 1
a4 = a3 – 1 = 1 – 1 = 0
a5 = a4 – 1 = 0 – 1 = -1
Thus, the first 5 terms of the sequence are 2, 2, 1, 0 and -1.
The corresponding series is
2 + 2 + 1 + 0 + (-1) + ……
14. The Fibonacci sequence is defined by
1 = a1 = a2 and an = an – 1 + an – 2, n > 2
Find an+1/an, for n = 1, 2, 3, 4, 5
Solution:
Given,
1 = a1 = a2
an = an – 1 + an – 2, n > 2
So,
a3 = a2 + a1 = 1 + 1 = 2
a4 = a3 + a2 = 2 + 1 = 3
a5 = a4 + a3 = 3 + 2 = 5
a6 = a5 + a4 = 5 + 3 = 8
Thus,
Exercise 9.2 Page No: 185
1. Find the sum of odd integers from 1 to 2001.
Solution:
The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.
It clearly forms a sequence in A.P.
Where, the first term, a = 1
Common difference, d = 2
Now,
a + (n -1)d = 2001
1 + (n-1)(2) = 2001
2n – 2 = 2000
2n = 2000 + 2 = 2002
n = 1001
We know,
Sn = n/2 [2a + (n-1)d]
Therefore, the sum of odd numbers from 1 to 2001 is 1002001.
2. Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Solution:
The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.
It clearly forms a sequence in A.P.
Where, the first term, a = 105
Common difference, d = 5
Now,
a + (n -1)d = 995
105 + (n – 1)(5) = 995
105 + 5n – 5 = 995
5n = 995 – 105 + 5 = 895
n = 895/5
n = 179
We know,
Sn = n/2 [2a + (n-1)d]
Therefore, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.
3. In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.
Solution:
Given,
The first term (a) of an A.P = 2
Let’s assume d be the common difference of the A.P.
So, the A.P. will be 2, 2 + d, 2 + 2d, 2 + 3d, …
Then,
Sum of first five terms = 10 + 10d
Sum of next five terms = 10 + 35d
From the question, we have
10 + 10d = ¼ (10 + 35d)
40 + 40d = 10 + 35d
30 = -5d
d = -6
a20 = a + (20 – 1)d = 2 + (19) (-6) = 2 – 114 = -112
Therefore, the 20th term of the A.P. is –112.
4. How many terms of the A.P. -6, -11/2, -5, …. are needed to give the sum –25?
Solution:
Let’s consider the sum of n terms of the given A.P. as –25.
We known that,
Sn = n/2 [2a + (n-1)d]
where n = number of terms, a = first term, and d = common difference
So here, a = –6
d = -11/2 + 6 = (-11 + 12)/ 2 = 1/2
Thus, we have
5. In an A.P., if pth term is 1/q and qth term is 1/p, prove that the sum of first pq terms is ½ (pq + 1) where p ≠ q.
Solution:
6. If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term
Solution:
Given A.P.,
25, 22, 19, …
Here,
First term, a = 25 and
Common difference, d = 22 – 25 = -3
Also given, sum of certain number of terms of the A.P. is 116
The number of terms be n
So, we have
Sn = n/2 [2a + (n-1)d] = 116
116 = n/2 [2(25) + (n-1)(-3)]
116 x 2 = n [50 – 3n + 3]
232 = n [53 – 3n]
232 = 53n – 3n2
3n2 – 53n + 232 = 0
3n2 – 24n – 29n+ 232 = 0
3n(n – 8) – 29(n – 8) = 0
(3n – 29) (n – 8) = 0
Hence,
n = 29/3 or n = 8
As n can only be an integral value, n = 8
Thus, 8th term is the last term of the A.P.
a8 = 25 + (8 – 1)(-3)
= 25 – 21
= 4
7. Find the sum to n terms of the A.P., whose kth term is 5k + 1.
Solution:
Given, the kth term of the A.P. is 5k + 1.
kth term = ak = a + (k – 1)d
And,
a + (k – 1)d = 5k + 1
a + kd – d = 5k + 1
On comparing the coefficient of k, we get d = 5
a – d = 1
a – 5 = 1
⇒ a = 6
8. If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference.
Solution:
We know that,
Sn = n/2 [2a + (n-1)d]
From the question we have,
On comparing the coefficients of n2 on both sides, we get
d/2 = q
Hence, d = 2q
Therefore, the common difference of the A.P. is 2q.
9. The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18th terms.
Solution:
Let a1, a2, and d1, d2 be the first terms and the common difference of the first and second arithmetic progression respectively.
Then, from the question we have
10. If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.
Solution:
Let’s take a and d to be the first term and the common difference of the A.P. respectively.
Then, it given that
Therefore, the sum of (p + q) terms of the A.P. is 0.
11. Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.
Prove that
Solution:
Let a1 and d be the first term and the common difference of the A.P. respectively.
Then according to the question, we have
Now, subtracting (2) from (1), we get
12. The ratio of the sums of m and n terms of an A.P. is m2: n2. Show that the ratio of mth and nth term is (2m – 1): (2n – 1).
Solution:
Let’s consider that a and b to be the first term and the common difference of the A.P. respectively.
Then from the question, we have
Hence, the given result is proved.
13. If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.
Solution:
Let’s consider a and b to be the first term and the common difference of the A.P. respectively.
am = a + (m – 1)d = 164 … (1)
We the sum of the terms is given by,
Sn = n/2 [2a + (n-1)d]
14. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Solution:
Let’s assume A1, A2, A3, A4, and A5 to be five numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 are in an A.P.
Here we have,
a = 8, b = 26, n = 7
So,
26 = 8 + (7 – 1) d
6d = 26 – 8 = 18
d = 3
Now,
A1 = a + d = 8 + 3 = 11
A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14
A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17
A4 = a + 4d = 8 + 4 × 3 = 8 + 12 = 20
A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23
Therefore, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.
15. If 
is the A.M. between a and b, then find the value of n.
Solution:
The A.M between a and b is given by, (a + b)/2
Then according to the question,
Thus, the value of n is 1.
16. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)th numbers is 5: 9. Find the value of m.
Solution:
Let’s consider a1, a2, … am be m numbers such that 1, a1, a2, … am, 31 is an A.P.
And here,
a = 1, b = 31, n = m + 2
So, 31 = 1 + (m + 2 – 1) (d)
30 = (m + 1) d
d = 30/ (m + 1) ……. (1)
Now,
a1 = a + d
a2 = a + 2d
a3 = a + 3d …
Hence, a7 = a + 7d
am–1 = a + (m – 1) d
According to the question, we have
Therefore, the value of m is 14.
17. A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs 5 every month, what amount he will pay in the 30th instalment?
Solution:
Given,
The first instalment of the loan is Rs 100.
The second instalment of the loan is Rs 105 and so on as the instalment increases by Rs 5 every month.
Thus, the amount that the man repays every month forms an A.P.
And the, A.P. is 100, 105, 110, …
Where, first term, a = 100
Common difference, d = 5
So, the 30th term in this A.P. will be
A30 = a + (30 – 1)d
= 100 + (29) (5)
= 100 + 145
= 245
Therefore, the amount to be paid in the 30th instalment will be Rs 245.
18. The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.
Solution:
It’s understood from the question that, the angles of the polygon will form an A.P. with common difference d = 5° and first term a = 120°.
And, we know that the sum of all angles of a polygon with n sides is 180° (n – 2).
Thus, we can say
Thus, a polygon having 9 and 16 sides will satisfy the condition in the question.
Exercise 9.3 Page No: 192
1. Find the 20th and nthterms of the G.P. 5/2, 5/4, 5/8, ………
Solution:
Given G.P. is 5/2, 5/4, 5/8, ………
Here, a = First term = 5/2
r = Common ratio = (5/4)/(5/2) = ½
Thus, the 20th term and nth term
2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Solution:
Given,
The common ratio of the G.P., r = 2
And, let a be the first term of the G.P.
Now,
a8 = ar 8–1 = ar7
ar7 = 192
a(2)7 = 192
a(2)7 = (2)6 (3)
3. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.
Solution:
Let’s take a to be the first term and r to be the common ratio of the G.P.
Then according to the question, we have
a5 = a r5–1 = a r4 = p … (i)
a8 = a r8–1 = a r7 = q … (ii)
a11 = a r11–1 = a r10 = s … (iii)
Dividing equation (ii) by (i), we get
4. The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7th term.
Solution:
Let’s consider a to be the first term and r to be the common ratio of the G.P.
Given, a = –3
And we know that,
an = arn–1
So, a4 = ar3 = (–3) r3
a2 = a r1 = (–3) r
Then from the question, we have
(–3) r3 = [(–3) r]2
⇒ –3r3 = 9 r2
⇒ r = –3
a7 = a r 7–1 = a r6 = (–3) (–3)6 = – (3)7 = –2187
Therefore, the seventh term of the G.P. is –2187.
5. Which term of the following sequences:
(a) 2, 2√2, 4,… is 128 ? (b) √3, 3, 3√3,… is 729 ?
(c) 1/3, 1/9, 1/27, … is 1/19683 ?
Solution:
(a) The given sequence, 2, 2√2, 4,…
We have,
a = 2 and r = 2√2/2 = √2
Taking the nth term of this sequence as 128, we have
Therefore, the 13th term of the given sequence is 128.
(ii) Given sequence, √3, 3, 3√3,…
We have,
a = √3 and r = 3/√3 = √3
Taking the nth term of this sequence to be 729, we have
Therefore, the 12th term of the given sequence is 729.
(iii) Given sequence, 1/3, 1/9, 1/27, …
a = 1/3 and r = (1/9)/(1/3) = 1/3
Taking the nth term of this sequence to be 1/19683, we have
Therefore, the 9th term of the given sequence is 1/19683.
6. For what values of x, the numbers -2/7, x, -7/2 are in G.P?
Solution:
The given numbers are -2/7, x, -7/2.
Common ratio = x/(-2/7) = -7x/2
Also, common ratio = (-7/2)/x = -7/2x
Therefore, for x = ± 1, the given numbers will be in G.P.
7. Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 …
Solution:
Given G.P., 0.15, 0.015, 0.00015, …
Here, a = 0.15 and r = 0.015/0.15 = 0.1
8. Find the sum to n terms in the geometric progression √7, √21, 3√7, ….
Solution:
The given G.P is √7, √21, 3√7, ….
Here,
a = √7 and
9. Find the sum to n terms in the geometric progression 1, -a, a2, -a3 …. (if a ≠ -1)
Solution:
The given G.P. is 1, -a, a2, -a3 ….
Here, the first term = a1 = 1
And the common ratio = r = – a
We know that,
10. Find the sum to n terms in the geometric progression x3, x5, x7, … (if x ≠ ±1 )
Solution:
Given G.P. is x3, x5, x7, …
Here, we have a = x3 and r = x5/x3 = x2
11. Evaluate: 
Solution:
12. The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.
Solution:
Let a/r, a, ar be the first three terms of the G.P.
a/r + a + ar = 39/10 …… (1)
(a/r) (a) (ar) = 1 …….. (2)
From (2), we have
a3 = 1
Hence, a = 1 [Considering real roots only]
Substituting the value of a in (1), we get
1/r + 1 + r = 39/10
(1 + r + r2)/r = 39/10
10 + 10r + 10r2 = 39r
10r2 – 29r + 10 = 0
10r2 – 25r – 4r + 10 = 0
5r(2r – 5) – 2(2r – 5) = 0
(5r – 2) (2r – 5) = 0
Thus,
r = 2/5 or 5/2
Therefore, the three terms of the G.P. are 5/2, 1 and 2/5.
13. How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?
Solution:
Given G.P. is 3, 32, 33, …
Let’s consider that n terms of this G.P. be required to obtain the sum of 120.
We know that,
Here, a = 3 and r = 3
Equating the exponents we get, n = 4
Therefore, four terms of the given G.P. are required to obtain the sum as 120.
14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
Solution:
Let’s assume the G.P. to be a, ar, ar2, ar3, …
Then according to the question, we have
a + ar + ar2 = 16 and ar3 + ar4 + ar5 = 128
a (1 + r + r2) = 16 … (1) and,
ar3(1 + r + r2) = 128 … (2)
Dividing equation (2) by (1), we get
r3 = 8
r = 2
Now, using r = 2 in (1), we get
a (1 + 2 + 4) = 16
a (7) = 16
a = 16/7
Now, the sum of terms is given as
15. Given a G.P. with a = 729 and 7th term 64, determine S7.
Solution:
Given,
a = 729 and a7 = 64
Let r be the common ratio of the G.P.
Then we know that, an = a rn–1
a7 = ar7–1 = (729)r6
⇒ 64 = 729 r6
r6 = 64/729
r6 = (2/3)6
r = 2/3
And, we know that
= 37 – 27
= 2187 – 128
= 2059
16. Find a G.P. for which sum of the first two terms is –4 and the fifth term is 4 times the third term.
Solution:
Consider a to be the first term and r to be the common ratio of the G.P.
Given, S2 = -4
Then, from the question we have
And,
a5 = 4 x a3
ar4 = 4ar2
r2 = 4
r = ± 2
Using the value of r in (1), we have
Therefore, the required G.P is
-4/3, -8/3, -16/3, …. Or 4, -8, 16, -32, ……
17. If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
Solution:
Let a be the first term and r be the common ratio of the G.P.
According to the given condition,
a4 = a r3 = x … (1)
a10 = a r9 = y … (2)
a16 = a r15 = z … (3)
On dividing (2) by (1), we get
18. Find the sum to n terms of the sequence, 8, 88, 888, 8888…
Solution:
Given sequence: 8, 88, 888, 8888…
This sequence is not a G.P.
But, it can be changed to G.P. by writing the terms as
Sn = 8 + 88 + 888 + 8888 + …………….. to n terms
19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2.
Solution:
The required sum = 2 x 128 + 4 x 32 + 8 x 8 + 16 x 2 + 32 x ½
= 64[4 + 2 + 1 + ½ + 1/22]
Now, it’s seen that
4, 2, 1, ½, 1/22 is a G.P.
With first term, a = 4
Common ratio, r =1/2
We know,
Therefore, the required sum = 64(31/4) = (16)(31) = 496
20. Show that the products of the corresponding terms of the sequences a, ar, ar2, …arn-1 and A, AR, AR2, … ARn-1 form a G.P, and find the common ratio.
Solution:
To be proved: The sequence, aA, arAR, ar2AR2, …arn–1ARn–1, forms a G.P.
Now, we have
Therefore, the above sequence forms a G.P. and the common ratio is rR.
21. Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
Solution:
Consider a to be the first term and r to be the common ratio of the G.P.
Then,
a1 = a, a2 = ar, a3 = ar2, a4 = ar3
From the question, we have
a3 = a1 + 9
ar2 = a + 9 … (i)
a2 = a4 + 18
ar = ar3 + 18 … (ii)
So, from (1) and (2), we get
a(r2 – 1) = 9 … (iii)
ar (1– r2) = 18 … (iv)
Now, dividing (4) by (3), we get
-r = 2
r = -2
On substituting the value of r in (i), we get
4a = a + 9
3a = 9
∴ a = 3
Therefore, the first four numbers of the G.P. are 3, 3(– 2), 3(–2)2, and 3(–2)3
i.e., 3¸–6, 12, and –24.
22. If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that aq-r br-p cp-q = 1
Solution:
Let’s take A to be the first term and R to be the common ratio of the G.P.
Then according to the question, we have
ARp–1 = a
ARq–1 = b
ARr–1 = c
Then,
aq–r br–p cp–q
= Aq–r × R(p–1) (q–r) × Ar–p × R(q–1) (r–p) × Ap–q × R(r –1)(p–q)
= Aq – r + r – p + p – q × R (pr – pr – q + r) + (rq – r + p – pq) + (pr – p – qr + q)
= A0 × R0
= 1
Hence proved.
23. If the first and the nth term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P2 = (ab)n.
Solution:
Given, the first term of the G.P is a and the last term is b.
Thus,
The G.P. is a, ar, ar2, ar3, … arn–1, where r is the common ratio.
Then,
b = arn–1 … (1)
P = Product of n terms
= (a) (ar) (ar2) … (arn–1)
= (a × a ×…a) (r × r2 × …rn–1)
= an r 1 + 2 +…(n–1) … (2)
Here, 1, 2, …(n – 1) is an A.P.
And, the product of n terms P is given by,
24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from 
.
Solution:
Let a be the first term and r be the common ratio of the G.P.
Since there are n terms from (n +1)th to (2n)th term,
Sum of terms from(n + 1)th to (2n)th term 
a n +1 = ar n + 1 – 1 = arn
Thus, required ratio = 
Thus, the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is 
.
25. If a, b, c and d are in G.P. show that (a2 + b2 + c2)(b2 + c2 + d2) = (ab + bc + cd)2.
Solution:
Given, a, b, c, d are in G.P.
So, we have
bc = ad … (1)
b2 = ac … (2)
c2 = bd … (3)
Taking the R.H.S. we have
R.H.S.
= (ab + bc + cd)2
= (ab + ad + cd)2 [Using (1)]
= [ab + d (a + c)]2
= a2b2 + 2abd (a + c) + d2 (a + c)2
= a2b2 +2a2bd + 2acbd + d2(a2 + 2ac + c2)
= a2b2 + 2a2c2 + 2b2c2 + d2a2 + 2d2b2 + d2c2 [Using (1) and (2)]
= a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2
= a2b2 + a2c2 + a2d2 + b2 × b2 + b2c2 + b2d2 + c2b2 + c2 × c2 + c2d2
[Using (2) and (3) and rearranging terms]
= a2(b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2+ c2 + d2)
= (a2 + b2 + c2) (b2 + c2 + d2)
= L.H.S.
Thus, L.H.S. = R.H.S.
Therefore,
(a2 + b2 + c2)(b2 + c2 + d2) = (ab + bc + cd)2
26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Solution:
Let’s assume G1 and G2 to be two numbers between 3 and 81 such that the series 3, G1, G2, 81 forms a G.P.
And let a be the first term and r be the common ratio of the G.P.
Now, we have the 1st term as 3 and the 4th term as 81.
81 = (3) (r)3
r3 = 27
∴ r = 3 (Taking real roots only)
For r = 3,
G1 = ar = (3) (3) = 9
G2 = ar2 = (3) (3)2 = 27
Therefore, the two numbers which can be inserted between 3 and 81 so that the resulting sequence becomes a G.P are 9 and 27.
27. Find the value of n so that 
may be the geometric mean between a and b.
Solution:
We know that,
The G. M. of a and b is given by √ab.
Then from the question, we have
By squaring both sides, we get
28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio
.
Solution:
Consider the two numbers be a and b.
Then, G.M. = √ab.
From the question, we have
29. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the
numbers are .
Solution:
Given that A and G are A.M. and G.M. between two positive numbers.
And, let these two positive numbers be a and b.
30. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?
Solution:
Given, the number of bacteria doubles every hour. Hence, the number of bacteria after every hour will form a G.P.
Here we have, a = 30 and r = 2
So, a3 = ar2 = (30) (2)2 = 120
Thus, the number of bacteria at the end of 2nd hour will be 120.
And, a5 = ar4 = (30) (2)4 = 480
The number of bacteria at the end of 4th hour will be 480.
an +1 = arn = (30) 2n
Therefore, the number of bacteria at the end of nth hour will be 30(2)n.
31. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Solution:
Given,
The amount deposited in the bank is Rs 500.
At the end of first year, amount = Rs 500(1 + 1/10) = Rs 500 (1.1)
At the end of 2nd year, amount = Rs 500 (1.1) (1.1)
At the end of 3rd year, amount = Rs 500 (1.1) (1.1) (1.1) and so on….
Therefore,
The amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times)
= Rs 500(1.1)10
32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Solution:
Let’s consider the roots of the quadratic equation to be a and b.
Then, we have
We know that,
A quadratic equation can be formed as,
x2 – x (Sum of roots) + (Product of roots) = 0
x2 – x (a + b) + (ab) = 0
x2 – 16x + 25 = 0 [Using (1) and (2)]
Therefore, the required quadratic equation is x2 – 16x + 25 = 0
Exercise 9.4 Page No: 196
Find the sum to n terms of each of the series in Exercises 1 to 7.
1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …
Solution:
Given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …
It’s seen that,
nth term, an = n ( n + 1)
Then, the sum of n terms of the series can be expressed as
2. 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
Solution:
Given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
It’s seen that,
nth term, an = n ( n + 1) ( n + 2)
= (n2 + n) (n + 2)
= n3 + 3n2 + 2n
Then, the sum of n terms of the series can be expressed as
3. 3 × 12 + 5 × 22 + 7 × 32 + …
Solution:
Given series is 3 ×12 + 5 × 22 + 7 × 32 + …
It’s seen that,
nth term, an = ( 2n + 1) n2 = 2n3 + n2
Then, the sum of n terms of the series can be expressed as
4. Find the sum to n terms of the series
Solution:
5. Find the sum to n terms of the series 52 + 62 + 72 + … + 202
Solution:
Given series is 52 + 62 + 72 + … + 202
It’s seen that,
nth term, an = ( n + 4)2 = n2 + 8n + 16
Then, the sum of n terms of the series can be expressed as
6. Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…
Solution:
Given series is 3 × 8 + 6 × 11 + 9 × 14 + …
It’s found out that,
an = (nth term of 3, 6, 9 …) × (nth term of 8, 11, 14, …)
= (3n) (3n + 5)
= 9n2 + 15n
Then, the sum of n terms of the series can be expressed as
7. Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + …
Solution:
Given series is 12 + (12 + 22) + (12 + 22 + 32 ) + …
Finding the nth term, we have
an = (12 + 22 + 32 +…….+ n2)
Now, the sum of n terms of the series can be expressed as
8. Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).
Solution:
Given,
an = n (n + 1) (n + 4) = n(n2 + 5n + 4) = n3 + 5n2 + 4n
Now, the sum of n terms of the series can be expressed as
9. Find the sum to n terms of the series whose nth terms is given by n2 + 2n
Solution:
Given,
nth term of the series as:
an = n2 + 2n
Then, the sum of n terms of the series can be expressed as
10. Find the sum to n terms of the series whose nth terms is given by (2n – 1)2
Solution:
Given,
nth term of the series as:
an = (2n – 1)2 = 4n2 – 4n + 1
Then, the sum of n terms of the series can be expressed as
Miscellaneous Exercise Page No: 199
1. Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.
Solution:
Let’s take a and d to be the first term and the common difference of the A.P. respectively.
We know that, the kth term of an A. P. is given by
ak = a + (k –1) d
So, am + n = a + (m + n –1) d
And, am – n = a + (m – n –1) d
am = a + (m –1) d
Thus,
am + n + am – n = a + (m + n –1) d + a + (m – n –1) d
= 2a + (m + n –1 + m – n –1) d
= 2a + (2m – 2) d
= 2a + 2 (m – 1) d
=2 [a + (m – 1) d]
= 2am
Therefore, the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term
2. If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.
Solution:
Let’s consider the three numbers in A.P. as a – d, a, and a + d.
Then, from the question we have
(a – d) + (a) + (a + d) = 24 … (i)
3a = 24
∴ a = 8
And,
(a – d) a (a + d) = 440 … (ii)
(8 – d) (8) (8 + d) = 440
(8 – d) (8 + d) = 55
64 – d2 = 55
d2 = 64 – 55 = 9
∴ d = ± 3
Thus,
When d = 3, the numbers are 5, 8, and 11 and
When d = –3, the numbers are 11, 8, and 5.
Therefore, the three numbers are 5, 8, and 11.
3. Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3 (S2– S1)
Solution:
Let’s take a and d to be the first term and the common difference of the A.P. respectively.
So, we have
4. Find the sum of all numbers between 200 and 400 which are divisible by 7.
Solution:
First let’s find the numbers between 200 and 400 which are divisible by 7.
The numbers are:
203, 210, 217, … 399
Here, the first term, a = 203
Last term, l = 399 and
Common difference, d = 7
Let’s consider the number of terms of the A.P. to be n.
Hence, an = 399 = a + (n –1) d
399 = 203 + (n –1) 7
7 (n –1) = 196
n –1 = 28
n = 29
Then, the sum of 29 terms of the A.P is given by:
Therefore, the required sum is 8729.
5. Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
Solution:
First let’s find the integers from 1 to 100, which are divisible by 2.
And, they are 2, 4, 6… 100.
Clearly, this forms an A.P. with the first term and common difference both equal to 2.
So, we have
100 = 2 + (n –1) 2
n = 50
Hence, the sum is
Now, the integers from 1 to 100, which are divisible by 5, are 5, 10… 100.
This also forms an A.P. with the first term and common difference both equal to 5.
So, we have
100 = 5 + (n –1) 5
5n = 100
n = 20
Hence, the sum is
Lastly, the integers which are divisible by both 2 and 5, are 10, 20, … 100.
And this also forms an A.P. with the first term and common difference both equal to 10.
So, we have
100 = 10 + (n –1) (10)
100 = 10n
n = 10
Thus, the required sum = 2550 + 1050 – 550 = 3050
Therefore, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.
6. Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.
Solution:
We have to first find the two-digit numbers, which when divided by 4, yield 1 as remainder.
They are: 13, 17, … 97.
As it’s seen that this series forms an A.P. with first term (a) 13 and common difference (d) 4.
Let n be the number of terms of the A.P.
We know that, the nth term of an A.P. is given by,
an = a + (n –1) d
So, 97 = 13 + (n –1) (4)
4 (n –1) = 84
n – 1 = 21
n = 22
Now, the sum of n terms of an A.P. is given by,
Therefore, the required sum is 1210.
7. If f is a function satisfying f(x + y) = f(x) f(y) for all x, y ∈ N such that , find the value of n.
Solution:
Given that,
f (x + y) = f (x) × f (y) for all x, y ∈ N … (1)
f (1) = 3
Taking x = y = 1 in (1), we have
f (1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9
Similarly,
f (1 + 1 + 1) = f (3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27
And, f (4) = f (1 + 3) = f (1) f (3) = 3 × 27 = 81
Thus, f (1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P. with the first term and common ratio both equal to 3.
We know that sum of terms in G.P is given by,
And it’s given that,
Hence, the sum of terms of the function is 120.
Therefore, the value of n is 4.
8. The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.
Solution:
Given that the sum of some terms in a G.P is 315.
Let the number of terms be n.
We know that, sum of terms is
Given that the first term a is 5 and common ratio r is 2.
Hence, the last term of the G.P = 6th term = ar6 – 1 = (5)(2)5 = (5)(32) = 160
Therefore, the last term of the G.P. is 160.
9. The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
Solution:
Let’s consider a and r to be the first term and the common ratio of the G.P. respectively.
Given, a = 1
a3 = ar2 = r2
a5 = ar4 = r4
Then, from the question we have
r2 + r4 = 90
r4 + r2 – 90 = 0
Therefore, the common ratio of the G.P. is ±3.
10. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.
Solution:
Let’s consider the three numbers in G.P. to be as a, ar, and ar2.
Then from the question, we have
a + ar + ar2 = 56
a (1 + r + r2) = 56
… (1)
Also, given
a – 1, ar – 7, ar2 – 21 forms an A.P.
So, (ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7)
ar – a – 6 = ar2 – ar – 14
ar2 – 2ar + a = 8
ar2 – ar – ar + a = 8
a(r2 + 1 – 2r) = 8
a (r – 1)2 = 8 … (2)
7(r2 – 2r + 1) = 1 + r + r2
7r2 – 14 r + 7 – 1 – r – r2 = 0
6r2 – 15r + 6 = 0
6r2 – 12r – 3r + 6 = 0
6r (r – 2) – 3 (r – 2) = 0
(6r – 3) (r – 2) = 0
r = 2, 1/2
When r = 2, a = 8
When r = ½, a = 32
Thus,
When r = 2, the three numbers in G.P. are 8, 16, and 32.
When r = 1/2, the three numbers in G.P. are 32, 16, and 8.
Therefore in either case, the required three numbers are 8, 16, and 32.
11. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.
Solution:
Let’s consider the terms in the G.P.to be T1, T2, T3, T4, … T2n.
The number of terms = 2n
Then, from the question we have
T1 + T2 + T3 + …+ T2n = 5 [T1 + T3 + … +T2n–1]
T1 + T2 + T3 + … + T2n – 5 [T1 + T3 + … + T2n–1] = 0
T2 + T4 + … + T2n = 4 [T1 + T3 + … + T2n–1] …… (1)
Now, let the terms in G.P. be a, ar, ar2, ar3, …
Then (1) becomes,
[Using sum of terms in G.P.]
ar = 4a
r = 4
Thus, the common ratio of the G.P. is 4.
12. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.
Solution:
Let’s consider the terms in A.P. to be a, a + d, a + 2d, a + 3d, … a + (n – 2) d, a + (n – 1)d.
From the question, we have
Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d
Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]
= 4a + (4n – 10) d
Then according to the given condition,
4a + 6d = 56
4(11) + 6d = 56 [Since a = 11 (given)]
6d = 12
d = 2
Hence, 4a + (4n –10) d = 112
4(11) + (4n – 10)2 = 112
(4n – 10)2 = 68
4n – 10 = 34
4n = 44
n = 11
Therefore, the number of terms of the A.P. is 11.
13. If
, then show that a, b, c and d are in G.P.
Solution:
Given,
On cross multiplying, we have
Also, given
On cross multiplying, we have
From (1) and (2), we get
b/a = c/b = d/c
Therefore, a, b, c and d are in G.P.
14. Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn
Solution:
Let the terms in G.P. be a, ar, ar2, ar3, … arn – 1…
Form the question, we have
Hence, P2 Rn = Sn
15. The pth, qth and rth terms of an A.P. are a, b, c respectively.
Show that (q – r) a + (r – p) b + (p -q) c = 0
Solution:
Let’s assume t and d to be the first term and the common difference of the A.P. respectively.
Then the nth term of the A.P. is given by, an = t + (n – 1) d
Thus,
ap = t + (p – 1) d = a … (1)
aq = t + (q – 1) d = b … (2)
ar = t + (r – 1) d = c … (3)
On subtracting equation (2) from (1), we get
(p – 1 – q + 1) d = a – b
(p – q) d = a – b
On subtracting equation (3) from (2), we get
(q – 1 – r + 1) d = b – c
(q – r) d = b – c
Equating both the values of d obtained in (4) and (5), we get
-a (q – r) – b (r – p) – c (p – q) = 0
a (q – r) + b (r – p) + c (p – q) = 0
Therefore, the given result is proved.
16. If a
are in A.P., prove that a, b, c are in A.P.
Solution:
17. If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P.
Solution:
Given, a, b, c,and d are in G.P.
So, we have
∴b2 = ac … (i)
c2 = bd … (ii)
ad = bc … (iii)
Required to prove (an + bn), (bn + cn), (cn + dn) are in G.P. i.e.,
(bn + cn)2 = (an + bn) (cn + dn)
Taking L.H.S.
(bn + cn)2 = b2n + 2bncn + c2n
= (b2)n+ 2bncn + (c2) n
= (ac)n + 2bncn + (bd)n [Using (i) and (ii)]
= an cn + bncn+ bn cn + bn dn
= an cn + bncn+ an dn + bn dn [Using (iii)]
= cn (an + bn) + dn (an + bn)
= (an + bn) (cn + dn)
= R.H.S.
Therefore, (an + bn), (bn + cn), and (cn + dn) are in G.P
– Hence proved.
18. If a and b are the roots of x2 – 3x + p = 0 and c, dare roots of x2 – 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q – p) = 17:15.
Solution:
Given, a and b are the roots of x2 – 3x + p = 0
So, we have a + b = 3 and ab = p … (i)
Also, c and d are the roots of x2 – 12x + q = 0
So, c + d = 12 and cd = q … (ii)
And given a, b, c, d are in G.P.
Let’s take a = x, b = xr, c = xr2, d = xr3
From (i) and (ii), we get
x + xr = 3
x (1 + r) = 3
And,
xr2 + xr3 =12
xr2 (1 + r) = 12
On dividing, we get
When r = 2, x = 3/(1 + 2) = 3/3 = 1
When r = -2, x = 3/(1 – 2) = 3/-1 = -3
Case I:
When r = 2 and x =1,
ab = x2r = 2
cd = x2r5 = 32
Case II:
When r = –2, x = –3,
ab = x2r = –18
cd = x2r5 = – 288
Therefore, in both the cases, we get (q + p): (q – p) = 17:15\
19. The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that 
.
Solution:
Let the two numbers be a and b.
A.M = (a + b)/ 2 and G.M. = √ab
From the question, we have
20. If a, b, c are in A.P,; b, c, d are in G.P and 1/c, 1/d, 1/e are in A.P. prove that a, c, e are in G.P.
Solution:
Given a, b, c are in A.P.
Hence, b – a = c – b … (1)
And, given that b, c, d are in G.P.
So, c2 = bd … (2)
Also, 1/c, 1/d, 1/e are in A.P.
So,
Now, required to prove that a, c, e are in G.P. i.e., c2 = ae
From (1), we have
2b = a + c
b = (a + c)/ 2
And from (2), we have
d = c2/ b
On substituting these values in (3), we get
Therefore, a, c, and e are in G.P.
21. Find the sum of the following series up to n terms:
(i) 5 + 55 + 555 + … (ii) .6 + .66 + . 666 + …
Solution:
(i) Given, 5 + 55 + 555 + …
Let Sn = 5 + 55 + 555 + ….. up to n terms
(ii) Given, .6 + .66 + . 666 + …
Let Sn = 06. + 0.66 + 0.666 + … up to n terms
22. Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + n terms.
Solution:
Given series is 2 × 4 + 4 × 6 + 6 × 8 + … n terms
∴ nth term = an = 2n × (2n + 2) = 4n2 + 4n
The 20th term,
a20 = 4 (20)2 + 4(20) = 4 (400) + 80 = 1600 + 80 = 1680
Therefore, the 20th term of the series is 1680.
23. Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + …
Solution:
The given series is 3 + 7 + 13 + 21 + 31 + …
S = 3 + 7 + 13 + 21 + 31 + …+ an–1 + an
S = 3 + 7 + 13 + 21 + …. + an – 2 + an – 1 + an
On subtracting both the equations, we get
S – S = [3 + (7 + 13 + 21 + 31 + …+ an–1 + an)] – [(3 + 7 + 13 + 21 + 31 + …+ an–1) + an]
S – S = 3 + [(7 – 3) + (13 – 7) + (21 – 13) + … + (an – an–1)] – an
0 = 3 + [4 + 6 + 8 + … (n –1) terms] – an
an = 3 + [4 + 6 + 8 + … (n –1) terms]
24. If S1, S2, S3 are the sum of first n natural numbers, their squares and their cubes, respectively, show that 9S22 = S3 (1 + 8S1).
Solution:
From the question, we have
Therefore, from (1) and (2), we have 9S22 = S3 (1 + 8S1).
25. Find the sum of the following series up to n terms:
Solution:
26. Show that
Solution:
nth term of the numerator = n(n + 1)2 = n3 + 2n2 + n
nth term of the denominator = n2(n + 1) = n3 + n2
27. A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual installments of Rs 500 plus 12% interest on the unpaid amount. How much will be the tractor cost him?
Solution:
Given, the farmer pays Rs 6000 in cash.
So, the unpaid amount = Rs 12000 – Rs 6000 = Rs 6000
From the question, the interest paid annually will be
12% of 6000, 12% of 5500, 12% of 5000, …, 12% of 500
Hence, the total interest to be paid = 12% of 6000 + 12% of 5500 + 12% of 5000 + … + 12% of 500
= 12% of (6000 + 5500 + 5000 + … + 500)
= 12% of (500 + 1000 + 1500 + … + 6000)
It’s seen that, the series 500, 1000, 1500 … 6000 is an A.P. with the first term and common difference both equal to 500.
Let’s take the number of terms of the A.P. to be n.
So, 6000 = 500 + (n – 1) 500
1 + (n – 1) = 12
n = 12
Now,
The sum of the A.P = 12/2 [2(500) + (12 – 1)(500)] = 6 [1000 + 5500] = 6(6500) = 39000
Hence, the total interest to be paid = 12% of (500 + 1000 + 1500 + … + 6000)
= 12% of 39000 = Rs 4680
Therefore, the tractor will cost the farmer = (Rs 12000 + Rs 4680) = Rs 16680
16680
28. Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual installment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?
Solution:
Given, Shamshad Ali buys a scooter for Rs 22000 and pays Rs 4000 in cash.
So, the unpaid amount = Rs 22000 – Rs 4000 = Rs 18000
Form the question, it’s understood that the interest paid annually is
10% of 18000, 10% of 17000, 10% of 16000 … 10% of 1000
Hence, the total interest to be paid = 10% of 18000 + 10% of 17000 + 10% of 16000 + … + 10% of 1000
= 10% of (18000 + 17000 + 16000 + … + 1000)
= 10% of (1000 + 2000 + 3000 + … + 18000)
It’s seen that, 1000, 2000, 3000 … 18000 forms an A.P. with first term and common difference both equal to 1000.
Let’s take the number of terms to be n.
So, 18000 = 1000 + (n – 1) (1000)
n = 18
Now, the sum of the A.P is given by:
Thus,
Total interest paid = 10% of (18000 + 17000 + 16000 + … + 1000)
= 10% of Rs 171000 = Rs 17100
Therefore, the cost of scooter = Rs 22000 + Rs 17100 = Rs 39100
29. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.
Solution:
It’s seen that,
The numbers of letters mailed forms a G.P.: 4, 42, … 48
Here, first term = 4 and common ratio = 4
And the number of terms = 8
The sum of n terms of a G.P. is given by:
Also, given that the cost to mail one letter is 50 paisa.
Hence, Cost of mailing 87380 letters = Rs 87380 x (50/100) = Rs 43690 = Rs 43690
Therefore, the amount spent when 8th set of letter is mailed will be Rs 43690.
30. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.
Solution:
Given, the man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.
Hence, the interest in first year = (5/100) x Rs 10000 = Rs 500
So, Amount in 15th year = Rs
= Rs 10000 + 14 × Rs 500
= Rs 10000 + Rs 7000
= Rs 17000
And, the amount after 20 years =
= Rs 10000 + 20 × Rs 500
= Rs 10000 + Rs 10000
= Rs 20000
Therefore, the amount in the 15th year is Rs 17000 and the total amount after 20 years will be Rs 20000.
31. A manufacturer reckons that the value of a machine, which costs him Rs 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
Solution:
Given, the cost of machine = Rs 15625
Also, given that the machine depreciates by 20% every year.
Hence, its value after every year is 80% of the original cost i.e., 4/5th of the original cost.
Therefore, the value at the end of 5 years =
= 5 × 1024 = 5120
Thus, the value of the machine at the end of 5 years will be Rs 5120.
32. 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
Solution:
Let’s assume x to be the number of days in which 150 workers finish the work.
Then from the question, we have
150x = 150 + 146 + 142 + …. (x + 8) terms
The series 150 + 146 + 142 + …. (x + 8) terms is an A.P.
With first term (a) = 150, common difference (d) = –4 and number of terms (n) = (x + 8)
Now, finding the sum of terms:
As x cannot be negative. [Number of days is always a positive quantity]
x = 17
Hence, the number of days in which the work should have been completed is 17.
But, due to the dropping out of workers the number of days in which the work is completed
= (17 + 8) = 25
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