Welcome to Class 11 Applied Maths Chapter 6, where we embark on an exciting journey into the world of advanced mathematical concepts tailored for Class 11 students.” Unlock the power of applied mathematics with expert solutions crafted by professionals at AppliedMath.com. Designed to propel students towards academic success, our meticulously curated ML Aggarwal Solutions for Applied Mathematics cater to Class 11 and class 12 students seeking mastery in their examinations. Every query from the CBSE ML Aggarwal Books finds a comprehensive answer on our platform, complete with detailed explanations and step-by-step solutions presented in an easily understandable language.
Dive into the world of applied mathematics and discover how our resources can elevate your understanding and performance. Keep reading to explore the wealth of ML Aggarwal Solutions for Class 11 and Class 12 Applied Mathematics.
Here we provide you with Class 11 Applied Maths Chapter 6, to help you gain a comprehensive understanding of the chapter and its concepts. https://appliedmathsolution.com/wp-admin/post.php?post=6&action=edit
Class 11 Applied Maths Chapter 6 Solutions
Sequences and Series
EXERCISE- 6.2
Q.1 Write the 10th term of each of the following sequences:
(i) -3, -2 1/4, -1 1/3, -3/4, 0, ……
(ii) √3, √12, √27, √48,…….
Ans.(i) a = -3
d =- 2 1/4 – (-3)
= -9/4 + 3/1
= (-9 + 12)4 = 3/4
Using, an = a + (n-1)d, we get,
a10 = a + 9d
= -3 + 9 x 3/4
= -3/1 + 27/4
= (-12 + 27)/4
= 15/4 = 3 3/4
(ii) a = √3
d = √12 – v3
= 2√3 – √3 = √3
Using, an = a + (n-1)d, we get,
a10 = a + 9d
= √3 + 9 x √3
= 10√3
= √100×3
= v300
Q.2 Find the 7th term from the end of the sequence -5, -2, 1, 4, …..,85.
Ans. New A.P. : 85, 82, 79,…..
a = 85, d = -3, n = 7
Using, an = a + (n-1)d, we get,
a7 = a + 6d
= 85 + 6 x (-3)
= 85 – 18 = 67
OR
7th term formed :
a(7 formed) = an – 6d
= 85 – 6 x 3
= 85 – 18 = 67
Q.3 Which term of the sequence 19, 91/5, 87/5,…. is the first negative term ?
Ans. Here, a = 19,
d = 91/5 – 19/1
= (91 – 95)/5 = -4/5
For first negative term, an < 0
Using, an = a + (n-1)d, we get,
19/1 + (n-1)(-4/5) < 0
[19 x 5 + (-4n + 4)]/5 < 0
95 – 4n + 4 < 0 x 5
99 – 4n < 0
99 < 4n
99/4 < n
24.75 < n
Therefore, n = 25
Q.4 Find the number of identical terms in the two sequences
1, 5, 9, 13, 17,…, 197 and 1, 4, 7, 10, 13, …, 196.
Ans. A.P 1 = 1, 5, 9, 13, 17,……19
a = 1, d = 4, an = 197
Using an = a + (n – 1)d, we get,
197 = 1 + (n – 1)4
197 – 1 = (n – 1)4
196 = (n – 1)4
196/4 = (n – 1)
49 = n – 1
n = 50
Also, an = 1 + (n – 1 )4
an = 1 + 4n – 4
an = 4n – 3
G.T. = 4n – 3
A.P 2 = 1, 4, 7,……196
a = 1, d = 3 , an = 196
Using an = a + (n – 1)d, we get,
196 = 1 + (n – 1)3
196 – 1 = (n – 1)3
195 = (n – 1)3
196/3 = (n – 1)
65 = n – 1
n = 66
G.T. = 1 + (n – 1)3
= 1 + 3n – 3
= 3n – 2
A.P 1 : a = 1, d = 4, an = 197, n = 50, G.T. = 4n – 3
A.P 2 : a = 1, d = 3, an = 196, n = 66, G.T. = 3n – 2
Let, pth term of A.P 1 be indentical with qth of A.P 2.
a + (p – 1)d = a + (q – 1)d
1 + (p – 1)4 = 1 + (q – 1)3
(p – 1)4 = (q – 1)3
(p – 1)/3 = (q – 1)/4 = k
p – 1 = 3k , q – 1 = 4k
p = 3k + 1, q = 4k + 1
Therefore, 3k + 1 < 50, 4k + 1 < 66
3k < 49 , 4k < 65
k < 49/3 , k < 65/4
k = 16.3 , k = 16.25
So, k = 16 terms , k = 16 terms
Q.5 If the sum of n terms of a sequence is given by Sn = 3n^2 – 2n, for all n ∊ N, find its 10th term.
Ans. an = Sn – [S(n – 1)]
= 3n^2 – 2n – [3(n-1)^2 – 2(n-1)]
= 3n^2 – 2n- [3(n^2 – 2n + 1) – 2n + 2]
= 3n^2 – 2n – 3n^2 + 6n – 3 + 2n – 2
an = 6n – 5
a10 = 6 x 10 – 5
= 55
Q.6 If the third term of an A.P. is 12 and the seventh term is 24, then find its 10th term.
Ans. a3 = 12 , a7 = 24
a + 2d = 12 , a + 6d = 24
By Elimination method,
a + 2d = 12
a + 6d = 24
(-) (-) (-)
-4d = -12
d = 3
From, a + 2d = 12
a + 2 x 3 = 12
a = 12 – 6
a = 6
a10 = a + 9d
= 6 + 9 x 3
= 33
Q.7 How many terms are there in the A.P. : -1, -5/6, -2/3, -1/2,…, -10/3 ?
Ans. Here, a = -1
d = -5/6 – (-1)
= -5/6 + 1/1
= (-5 + 6)/1 = 1/6
an = 10/3
n = ?
Using, an = a + (n – 1)d, we get,
10/3 = -1 + (n – 1)(1/6)
10/3 + 1/1 = (n – 1)(1/6)
(10 + 3)/3 = (n – 1)(1/6)
13/3 x 6/1 = n – 1
26 = n – 1
n = 27
Q.8 If x, y and z are in A.P., then show that (x + 2y – z) (x + z – y) (z + 2y – x) = 4 xyz.
Ans. x, y, z = A.P
we know that, 2A2 = A1 + A3.
2y = x + z…….(eq. i)
L.H.S.
(x + 2y – z)(x + z – y)(z + 2y – x)
From (eq.i)
(x + x + z – z)(2y – y)((z + x + z – x)
2x X y X 2z
4xyz = R.H.S
Q.9 If x, 2x + 1 and 4x – 1 are first three terms of an A.P., then find the fourth term of the A.P.
Ans. A.P : x, 2x + 1, 4x – 1
We know that in an A.P,
A2 – A1 = A3 – A2
2x + 1 – x = 4x – 1 -(2x + 1)
x + 1 = 4x – 1 – 2x – 1
x + 1 = 2x – 2
0 = 2x – x – 2 – 1
0 = x – 3
x = 3
Now, A.P is 3, 7, 11……..
a = 3, d = 4
a4 = a3 + d
a4 = 11 + 4
= 15
Q.10 The first and the last terms of an A.P. are 1 and 11. If the sum of the terms is 36, then find the number of terms.
Ans. A1 = 1 , An = 11, Sn = 36, n = ?
Using, Sn = n/2[A1 + An], we get,
36 = n/2[1 + 1]
36 x 2 = n x 12
(36 x 2)/12 = n
3 x 2 = n
n = 6.
Q.11 Find the sum of the following series:
(i) 4 + 7 + 10 +…. to 100 terms
(ii) 1 + 4/3 + 5/3 + 2…… to 19 terms.
(iii) 0.5 + 0.51 + 0.52… to 1000 terms.
Ans. (i) 4 + 7 + 10 +…. to 100 terms
Here, A1 = 4, d = 3, n = 100
Using, Sn = n/2[2a + (n – 1)d], we get,
S100 = 100/2[2 x 4 + 99 x 3]
= 50[8 + 297]
= 50 x 305
= 15250
(ii) 1 + 4/3 + 5/3 + 2…… to 19 terms.
Here, A1 = 1, d = 1/3, n = 19
Using, Sn = n/2[2a + (n – 1)d], we get,
S19 = 19/2[2 x 1 + 18 x 1/3]
= 19/2[2 + 6]
= 19/2 x 8
= 76
(iii) 0.5 + 0.51 + 0.52… to 1000 terms.
Here, A1 = 0.5, d = 0.01, n = 1000
Using, Sn = n/2[2a + (n – 1)d], we get,
S1000 = 1000/2[2 x 0.5 + 999 x 0.01]
= 500[1.0 + 99.9]
= 500 x 10.99
= 5495
Q.12 In a sequence, the sum of first n terms is pn + qn^2 where p, q are constants. Show that the sequence is an A.P. and find the common difference.
Ans. an = Sn – [S(n-1)]
= pn + qn^2 – [p(n-1) + q(n-1)^2]
= pn + qn^2 – pn + p – qn^2 + 2qn – q
= 2qn + p – q
a1 = 2q + p – q = q + p
a2 = 2q(2) + p – q = 3q + p
a3 = 2q(3) + p – q = 5q + p
d = 3q + p – (q + p) = 2q
d = 5q + p – (3q + p) = 2q
Therefore, it es an A.P
d = 2q
Q.13 Show that the sum of (m + n)Th and (m – n)Th terms of an A.P. is equal to twice the mth term.
Ans. A(m+n) + A(m-n) = 2Am
L.H.S,
Using, An = a + (n – 1)d, we get,
A + (m + n – 1)d + A + (m – n – 1)d
A + dm + dn – d + A + dm – dn – d
2A + 2dm – 2d
2(A + dm – d)
2[A + (m – 1)d]
2Am = R.H.S
Q.14 In an A.P., if mth term is n and nth term is m, where m ≠ n, find
(i) the pth term (ii) (m + n)Th term.
Ans. Am = n, An = m
Using, an = a + (n – 1)d, we get,
Am = A + (m – 1)d = n
A + dm – d = n…….(i)
An = A + (n – 1)d = m
A + dn – d = m……..(ii)
Subtract (i) & (ii), we get,
A + dm – d = n
A + dn – d = m
(-) (-) (+) (-)
dm – dn + 0 = n – m
-d(n – m) = n – m
-d = (n-m)/(n-m) = 1
Therefore, d = -1
Subtract value of d in eq (i), we get
A (-1)m – (-1) = n
A – m + 1 = n
A = n + m – 1
Now, A = n + m – 1, d = -1
(i) Ap = n + m – 1 +(p-1)(-1)
= n + m – 1 – p + 1
= n + m – p
(ii) A(m+n) = (n + m – 1) + (m + n – 1) (-1)
= (m + n – 1) – (m + n – 1) = 0
15. If a, b, c are pth, qth and rth terms respectively of an A.P., prove that
(i) p(b – c) + q(c – a) + r(a – b)=0
(ii) a(q – r) + b(r – p) + c(p – q)=0.
Ans. Ap = a, Aq = b, Ar = c
Using, An = A + (n-1)d, we get,
A + (p-1)d = a
A + dp – d = a
dp = a – A + d
p = (a – A + d)/d
q = (b – A + d)/d
r = (c – A +d)/d
(i) p(b – c) + q(c – a) + r(a – b)=0
(a – A + d)/d(b-c) + (b – A + d)/d(c-a) + (c – A + d)/d(a-b)
(ab – Ab + db – ac + Ac – cd + bc – Ac + cd – ab + Aa – ad + ac – Aa + ad – bc + Ab – bd)/d
0/d = 0
(ii) a(q – r) + b(r – p) + c(p – q)=0.
(A + dp – d)(q-r) + (A + dq – d)(r-p) + (A + dr – d)(p-q)
Aq + dpq – dq – Ar + dpr + dr + Ar + dqr – dr – Ap – dpq + dp + Ap + dpr – dp – Aq – dqr + dq
= 0
Q.16 If the sum of n terms of an A.P. is 3n^2+5n and its mth term is 164, find the value of m.
Ans. am = Sm – S(m-1) = (3m^2 + 5m) – [3(m-1)^2 + 5(m-1)].
164 = 3m^2 + 5m – [3(m^2 + 1^2 – 2m) + 5m – 5]
164 = 3m^2 + 5m – [3m^2 + 3 – 6m + 5m – 5]
164 = 3m^2 + 5m -[3m^2 – m – 2]
164 = 3m^2 + 5m – 3m^2 + m + 2
164 – 2 = 6m
162 = 6m
162/6 = m
m = 27
Q.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?
Ans. A.P : 100, 105, 110,….
a = 100, d = 5, n = 30
Using, an = a + (n-1)d, we get,
a30 = 100 + 29 x 5
= 100 + 145
= Rs.245
Q.18 How many terms of the A.P. 17, 15, 13, … must be taken so that the sum is 72? Explain the double answer.
Ans. A.P : 17, 15, 13,……….
a = 17, d = -2, Sn = 72
Using, Sn = n/2[2a + (n-1)d]
72 = n/2[2 x 17 + (n-1)(-2)]
144 = n[34 – 2n + 2]
144 = n[36 – 2n]
144 = 36n – 2n^2
2n^2 – 36n + 144 = 0
Divide the eq. by 2, we get,
n^2 – 18n + 72 = 0
n^2 – 12n – 6n + 72 = 0
n(n – 12) – 6(n – 12) = 0
(n – 12) (n – 6) = 0
Therefore,
n = 6 or 12.
Q.19 In an A.P., if the 12th term is -13 and the sum of first four terms is 24, what is the sum of first 10 terms?
Ans. A12 = -13
a + 11d = -13
a = -13 – 11d……….(i)
S4 = 24
4/2[2a + 3d] = 24
2[2a + 3d] = 24
2a + 3d = 12
From (i), we get,
2(-13-11d) + 3d = 12
-26 – 22d + 3d = 12
-19d = 26 +12
-19d = 38
d = 38/-19 = -2
Substitute value of d in eq. (i)
a = -13 -11 x -2
= -13 + 22 = 9
Using, Sn = n/2[2a + (n-1)d], we get,
S10 = 10/2[2 x 9 + 9 x -2]
= 5 [ 18 – 18]
= 5 x 0 = 0
Q.20 Find the sum of all natural numbers lying between 100 and 1000 which are multiples of 5.
Ans. A.P = 105, 110, …….,995
A1 = 105, An = 995, d = 5, n = ?
Using, An = A + (n-1)d, we get,
995 = 105 + (n-1)5
995 – 105 = (n-1)5
890/5 = n – 1
178 = n – 1
n = 179
Now, using Sn = n/2[A1 + An], we get,
S179 = 179/2 [105 + 995]
= 179/2 x 1100
= 179 x 550
= 98450
Q.21 Find the sum of integers from 100 to 500 that are divisible by 2 and 3.
Ans. 102, 108,…….498.
Clearly, they form an A.P.
Using, an = a + (n-1)d
498 = 102 + (n-1)6
498 – 102 = (n-1)6
396/6 = n – 1
66 + 1 = n
n = 67
a = 102, n = 67, an = 498
Using, Sn = n/2[a + an]
S67 = 67/2[102 + 498]
= 67/2 x 600
= 67 x 300
= 20100
Q.22 In an A.P., if pth term is 1/p and qth term id 1/q, the prove that the sum of first pq terms is 1/2(pq + 1) where p ≠ q.
Ans. ap = 1/q, aq = 1/p
Using, an = a + (n-1)d, we get,
ap = a + (p-1)d = 1/q
a + dp – d = 1/q………(i)
similarly, aq = a + (q-1)d = 1/p
a + dq – d = 1/p…….(ii)
From (i) & (ii)
a + dp – d = 1/q
a + dq – d = 1/p
(-) (-) (+) (-)
0 + dp – dq + 0 =1/q – 1/p
d(p-q) = (p-q)/pq
d = (p-q)/pq x 1/(p-q)
Therefore, d = 1/pq
putting value of d in eq. (i)
a + (p-q)d = 1/q
a + p x 1/pq – 1 x 1/pq = 1/q
a + 1/q – 1/pq = 1/q
a = 1/q – 1/q + 1/pq
a = 1/pq
Now, Using Sn = n/2[2a + (n-1)d]
Spq = pq/2 [2 x 1/pq + (pq – 1)1/pq]
Spq = pq/2 [2/pq + pq x 1/pq – 1 x 1/pq]
= pq/2 [2/pq +1 – 1/pq]
= pq/2 [2/pq + 1/pq + 1]
= pq/2 [1/pq + 1/1]
= pq/2 [(1+pq)/pq]
= 1/2 [pq + 1]
Hence proved.
Q.23 The sum of the first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term is 1/3. Calculate the first and the thirteenth term.
Ans. S6 = 42, a10/a30 = 1/3
Using, Sn =n/2[2a + (n-1)d], we get,
6/2[2a + 5d] = 42
2a – 5d = 42/3
2a + 5d = 14……….(i)
Using, an = a + (n-1)d ,we get,
(a+9d)/(a+29d) = 1/3
3(a+9d) = 1(a+29d)
3a + 27d = a + 29d
3a – a = 29d – 27d
2a = 2d
Therefore, a = d
from eq.(i)
2a + 5a = 14
7a = 14
a = 14
So, d = 14
Now, we know that
a13 = a + 12d
= 2 + 12 x 2
= 2 + 24 = 26
Q.24 If the sums of n terms of two arithmetic progression are in the ratio (3n+8): (7n +15), find the ratio of their 12th terms.
Ans. Sum of first A.P/Sum of second A.P = (3n+8)/(7n+15)
{n/2[2a+(n-1)d]}/{n/2[2A+(n-1)D} = (3n+8)/(7n+15)
Dividing Numerator & Denominator by 2
{a+[(n-1)/2]d}/{A+[(n-1)/2]d} = (3n+8)/(7n+15)
Now, put n = 23,
{a+[(23-1)/2]d}/{a+[(23-1)/2]d} = (3×23+8)/(7×23+15)
(a + 11d)/(A + 11D) = (69 + 8)/(161 + 15)
a12/A12 = 77/176
= 7/16
Q.25 If the ratio of the sum of m terms of an A.P. to the sum of n terms is m^2:n^2, show that ratio of the pth and qth term is (2p-1): (2q-1).
Ans. Sm/Sn = m^2/n^2
Using, Sn = n/2[2a + (n-1)d], we get,
m/2[2a(m-1)d/n/2[2a(n-1)d] = m^2/n^2
[2a+(m-1)d]/[2a+(n-1)d] = m/n
{a+[(m-1)/2]d}/{a+[(n-1)/2]} = m/n
put, m = 2p – 1 & n = 2q – 1
{a+[(2p-1-1)/2]d}/{a+[(2q-1-1)/2]d} = (2p-1)/(2q-1)
{a+[(2p-2)/2]d}/{a-[2q-2)/2]d} = (2p-1)/(2q-1)
{a+[2(p-1)/2]d}/{a+[2(q-1)/2]d} = (2p-1)/(2q-1)
[a+(p-1)d]/[a+(q-1)d] = (2p-1)/(2q-1)
Q.26 (i) If a + b + c ≠ 0 and (b + c)/,(c + a)/b, (a + b)/c are in A.P., prove
that 1/a, 1/b, 1/c are also in A.P.
(ii) show that a^2, b^2, c^2 are in A.P. iff 1/(b + c), 1/(c + a), 1/(a + b) are in A.P.
Ans. (i) Given : (b + c)/a, (c + a)/b, (a + b)/c
Add 1 to each term of A.P
(b + c)/a + 1/1, (c + a)/b + 1/1, (a + b)/c + 1/1
(b + c + a)/2, (c + a + b)/2, (a + b + c)/2
Dividing the three terms of A.P by a + b + c, we get,
(b + c + a)/2(a + b + c), (c + a + b)/2(a + b + c), (a + b + c)/2(a + b + c)
Therefore, 1/a, 1/b, 1/c are in A.P
(ii) Given : 1/(b + c), 1/(c + a), 1/(a + b)
2a2 = a1 + a3
2/(c + a) = 1/(b + c) + 1/(a + b)
2/(c + a) = (a + b + b + c)/(b + c)(a + b)
2/(c + a) = (a + 2b + c)/(ab + ac + b^2 + bc)
2ab + 2ac + 2b^2 + 2bc = ac + 2bc + c^2 + a^2 + 2ab + ac
2b^2 = a^2 + c^2
b^2 + b^2 = a^2 + c^2
b^2 – a^2 = c^2 – b^2
Therefore, a^2, b^2, c^2 are in A.P.
Q.27 (i) Find three number in A.P. whose sum is 21 and the product is 315.
(ii) The sum of three consecutive numbers in an A.P. is 24 and the sum of their squares is 194. Find the numbers.
(iii) The sum of three numbers in an A.P. is 30, and the ratio of first to third is 3:7. Find the numbers.
Ans. (i) Let, the three no. in A.P be
(a – d), a, (a + d)
Now, according to the question,
(a – d) + a + (a + d) = 21
a – d + a + a + d = 21
3a = 21
a = 7
(7 – d), 7, (7 + d)
Also, (7 – d) x 7 x (7 + d) = 315
(7 – d) (7 + d) = 315/7
7^2 – d^ 2 = 45
49 – 45 = d^2
4 = d^2
d = ±2
when, d = 2
values = 5, 7, 9
when, d = -2
values = 9, 7, 5
(ii) Let, the three consecutive terms of A.P be,
(a – d), a, (a + d)
[a – d] + a + [a + d] = 24
3a = 24
a = 8
(8 – d), 8, (8 + d)
(8 – d)^2 + 8^2 + (8 + d)^2 = 194
8^2 + d^2 – 16d + 8^2 + 8^2 + d^2 + 16d = 194
64 x 3 + 2d^2 = 194
192 + 2d^2 = 194
2d^2 = 194 – 192
2d^2 = 2
d = ±1
when, d = 1
values = 7, 8, 9
when, d = -1
values = 9, 8, 7
(iii) Let, the three terms of A.P be,
(a – d), a, (a + d)
a – d + a + a + d = 30
3a = 30
a = 10
(10 – d), 10, (10 + d)
According to the question,
(10-d)/(10+d) = 3/7
70 – 7d = 30 + 3d
70 – 30 = 3d + 7d
40 = 10d
d = 4
Numbers = 6, 10, 14
Q.28 The income of a person is Rs.300000 in the first year and he receives an increase of Rs.10000 to his income per year for the next 19 years. Find the total amount he received in 20 years.
Ans. Income in first year = Rs.300000
Income in second year = Rs.310000
Since, income is increased by a fixed value, they form an A.P
Here, a = 300000, d = 10000, n = 20
Using, Sn = n/2[2a + (n-1)d], we get,
S20 = 20/2[2 x 300000 + 19 x 10000]
= 10[600000 + 190000]
= 10 x 790000
= Rs.7900000
Q.29 A person was drawing a monthly salary of Rs.25000 in 11th year of service and a salary of Rs.29000 in the 19th year. Given that pension is half the salary at retirement time, find his monthly pension if he had put in 25 years of service before retirement. Assume that annual increment is constant.
Ans. a11 = 25000, a19 = 29000
a + 10d = 25000……….(i)
a + 18d = 29000……..(ii)
Subtract (i) from (ii), we get
a + 18d = 29000
a + 10d = 25000
(-) (-) (-)
8d = 4000
d = 4000/8 = 500
From (i), we egt,
a + 10 x 500 = 25000
a = 25000 – 5000
a = 20000
Salary after 25 years,
a25 = a + 24d
= 20000 + 24 x 500
= 20000 + 12000
= 32000
Pension = 1/2 x 32000
= Rs.16000
Q.30 A man saved Rs.66000 in 20 years. In each succeeding year after the first year he saved Rs.200 than what he saved in the previous year. How much did he save in the first year?
Ans. S20 = 66000, n = 20, d = 200, a = ?
Using, Sn = n/2[2a + (n – 1)d]
S20 = 20/2[2a + 19 x 200]
66000 = 10 [2a + 3800]
66000/10 = 2a + 3800
6600 – 3800 = 2a
2800 = 2a
a = 2800/2
a = 1400
Q.31 A man accepts a position with an initial salary of Rs.5200 per month. It is understood that he will receive an automatic increase of Rs.320 in the very next month and each month thereafter.
(i) Find his salary for the tenth month.
(ii) What is his total earnings during the first year?
Ans. (i) a = Rs.5200, d = 320, n = 10
Using, an = a + (n-1)d, we get,
a10 = 5200 + 9 x 320
= 5200 + 2880
= Rs.8080
(ii) Using, Sn = n/2[2a + (n-1)d]
S12 = 12/2[2 x 5200 + 11 x 320]
= 6[10400 + 3520]
= 6 x 13920
= 83520
Q.32 A person buys every year National Savings Certificates of value exceeding the last year’s purchase by Rs.5000. After 10 years, he finds the total face value of certificates purchased by him is Rs.275000. Find the value of the certificates purchased by him in the first year.
Ans. S10 = 275000, d = 5000, n = 10
Using, Sn = n/2[2a + (n-1)d]
275000 = 10/2[2a + 9 x 5000]
275000 = 5[2a + 45000]
275000/5 = 2a + 45000
55000 – 45000 = 2a
10000 = 2a
a = 5000
Q.33 An N.G.O. wants to invest Rs.300000 in a bond which pays 5% annual interest first year, 6% interest second year, 7% interest third year and so on. The interest received per year is to be utilized for the education of poor people. If the investment is done for 10 years, find the total interest received by N.G.O.
Ans. Interest for 1 year = 300000 x 5/100
= Rs.15000
Interest for 2 year = 300000 x 6/100
= Rs.18000
Interest for 3 year = 300000 x 7/100
= Rs.21000
Clearly, the interest amount forms an A.P
Here, a = 15000, d = 3000, n = 10
For finding out total interest we would use concept of sum of n terms.
Using, Sn = n/2[2a + (n-1)d]
S10 = 10/2[2 x 15000 + 9 x 3000]
= 5[30000 + 27000]
= 5 x 57000
= Rs.285000
Q.34 Kumar buys a Maruti car for Rs.2,40,000 at a down payment of ₹1,20,000 and the rest in annual instalments of Rs.10,000 plus 12% interest on the unpaid amount. Eventually what will the car cost him?
Ans. Original Value = 240000
Down payment = 120000
Loan = 120000
EMI 1 : 10000 + 12/100 X 120000
10000 + 14400 = 24400
EMI 2 : 10000 + 12/100 X 110000
10000 + 13200 = 23200
EMI 1 : 10000 + 12/100 X 100000
10000 + 12000 = 22000
Clearly, the EMI forms an A.P
A.P = 24400, 23200, 22000…..
a = 24400, d = -1200, n = 12
Using, Sn = n/2[2a + (n-1)d], we get,
S12 = 12/2[2 x 24400 + 11 x -1200]
= 6[48800 – 13200]
= 6 x 35600
= 213600
Cost = 120000 + 213600
= Rs.333600
FAQ’s related to Class 11 Applied Maths Chapter 6 on Sequences and Series:
Q.1 What is a sequence?
Ans. A sequence is an ordered list of numbers following a particular pattern. Each number in a sequence is called a term.
Q.2 What is a series?
Ans. A series is the sum of the terms of a sequence. If we add the terms of a sequence, we get a series.
Q.3 What is the difference between an arithmetic sequence and a geometric sequence?
Ans.
- Geometric Sequence – A geometric sequence (or geometric progression) is a sequence in which each term is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio (r).
- Arithmetic Sequence – An arithmetic sequence (or arithmetic progression) is a sequence in which the difference between consecutive terms is constant. This difference is called the common difference (d).
These are a few Frequently Asked Questions relating to Class 11 Applied Maths Chapter 6
In Class 11 Applied Maths chapter 6, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Class 11 Applied Maths Chapter 6 promises an enriching learning experience that will set you on the path to success. Class 11 Applied Maths Chapter 6, we delve deep into advanced mathematical concepts that are crucial for understanding.
Class 11 Applied Maths Chapter 6 Exercise :
