In this page, you can find the complete solutions of the first exercise of Sequence and Series chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan.

In the above-mentioned book, sequence and series is the 4th chapter and has two exercises only. Out of which, this is the solution of the first exercise.

Disclaimer:

Answers mentioned here are not solved by teachers. These are the solutions written by a student of Grade 11. Answers are all correct. However, the language or process of solving the questions might be informal and in examinations, you might have to add little more language and show more calculations than what has been done here. So, we highly encourage you to view these solutions as guide rather than just copying everything mentioned here. Few questions have been typed while most of them have been updated as pictures.

1 a) If the 4th term of an AP is zero, show that the 8th term is double of the 6th term.

Solution:
We know,
$t_n = a+(n-1)d$
Given,
$t_4 = 0$
$or, a + 3d = 0$
$or, a = - 3d$
Now,
$t_6 = a + 5d = -3d + 5d = 2d$

And,
$t_8 = a + 7d = -3d + 7d = 4d$

$or, t_8 = 4d = 2 * 2d$
$\therefore t_8 = 2*t_6$
Hence, it is proved that the 8th term of the given AP is double of the 6th term.



1 b) If the mth term of an AP is n and nth term is m, show that the pth term is m+n-p.

Solution:
We know,
$t_n = a + (n-1)d$
So,
$t_m = a + (m-1)d = n$
$or, a = n - (m-1)d$ --- (i)
And,
$t_n = a + (n-1)d = m$
$or, a = m - (n-1)d$ --- (ii)
Comparing values of a from equation (i) and (ii), we get,

$n - (m-1)d = m - (n-1)d$
$or, n - md + d = m - nd +d$
$or, m - n = - (m -n)d$
$\therefore d = -1$
Put value of d = -1 in equation (i), we get, a = (m+n-1)

Now,
$t_p = a + (p-1)d = (m+n-1) + (p-1)(-1)$

$or, t_p = m + n - 1 - p + 1$
$\therefore t_p = m + n - p$ proved


1 c) If the pth, qth and rth term of an AP is a,b, and c respectively, prove that:
$a(q-r) + b(r-p) + c(p-q) = 0$

Solution:
Given,
$t_p = A + (p-1)d = a$ --- (i)

$t_q = A + (q-1)d = b$ --- (ii)

$t_r = A + (r-1)d = c$ --- (iii)

To prove: $a(q-r) + b(r-p) + c(p-q) = 0$

Subtracting equation (ii) from (iii), we get,

$(q-r)d = b -c \implies (q - r) = \dfrac{b-c}{d}$

Subtracting equation (iii) from (i), we get,

$(r-p)d = c - a \implies (r-p) = \dfrac{c-a}{d}$

Subtracting equation (i) from (ii), we get,

$(p-q)d = a -b \implies (p-q) = \dfrac{a-b}{d}$

Put values in LHS of required proof, we get,

LHS = $a(q-r) + b(r-p) + c(p-q) = 0$

$= \dfrac{a(b-c)}{d} + \dfrac{b(c-a)}{d} + \dfrac{c(a-b)}{d}$

$= \dfrac{ab -ac + bc - ab + ac - bc}{d}$

$= \dfrac{0}{ab}$
$= 0$
= RHS
Hence,
The required relation is proved.



2 a) If the 5th term of an AP is twice the 7th term, show that the sum of the 17 terms is zero.

Solution:
We know,
$t_n = a + (n-1)d$
Given,
$t_5 = 2(t_7)$
$or, a + 4d = 2 ( a + 6d)$
$or, a + 4d = 2 a + 12d$
$or, - 8d = a \implies a = -8d$ --- (i)

Now,
$S_n = \dfrac{n}{2} [2a + (n-1)d]$

$or, S_{17} = \dfrac{17}{2} [ 2(-8d) + 16d]$

$or, S-{17} = \dfrac{17}{2} [-16d+16d]$

$\therefore S_{17} = 0$
Hence,
The required sum of the 17 terms is zero is proved.


2 b) If the sum of the p terms of an AP is equal tot he sum of the q terms, show that the sum of the (p+q) terms is zero.

Solution:

We know,
$S_n = \dfrac{n}{2} [2a + (n-1)d]$

Given,
$S_p = S_q$
$or, \dfrac{p}{2} [2a + (p-1)d] = \dfrac{q}{2} [2a + (p-1)d]$

$or, 2ap + (p-1)pd = 2aq + (q-1)qd$

$or, 2a(p -q) = (q-1)qd - (p-1)pd$

$or, 2a (p-q) = d { q(q-1) - p(p-1)}$

$or, 2a (p-q) = d {q^2 - q - p^2 + p}$

$or, 2a (p-q) = -d {p^2 - q^2 - p + q}$

$or, 2a ( p-q) = -d {(p+q)(p-q) - (p -q)}$

$or, 2a (p-q) = -d (p-q) [p + q -1]$

$or, -2a = d [p + q - 1]$ --- (i)

Now,

$S_{p+q} = \dfrac{p+q}{2} [2a + [p+q -1]d]$

$or, S_{p+q} = \dfrac{p+q}{2} [2a - 2a]$

$\therefore S_{p+q} = 0$
Hence, it is proved that the sum of the (p+q) terms of the given arithmetic series is 0.


Solutions to questions from 3 to 6 are mentioned in the form of pictures.



















7 a) Find the two numbers whose arithmetic mean is 25 and geometric mean is 20.

Solution:

Let the two unknown numbers be a and b.

Given,
AM is 25
$or, \dfrac{a+b}{2} = 25 \implies (a+b) = 50$

$or, a = 50 - b$ --- (i)
And,
GM is 20
$or, \sqrt{ab} = 20$
$or, ab = 20^2 \implies ab = 400$

$or, a = \dfrac{400}{b}$ --- (ii)

Put value of a from equation (i) in equation (ii), we get,

$or, 50 -  b = \dfrac{400}{b}$

$or, b(50 -b) = 400$
$or, b^2 - 50 b + 400 = 0$
$or, (b - 10)(b -40) = 0$
Either,
$(b - 10) = 0 \implies b = 10$
So, $ a= 40$
Or,
$(b - 40) = 0 \implies b = 40$
So, $a = 10$
Hence, the required unknown numbers (a,b) are either (10,40) or (40,10). 


7 b) The AM inserted between two numbers exceeds their GM by 2 and Gm exceeds the HM by 16. Find the numbers.

Solution:
Let the two numbers be a and b.

We have,
$AM = \dfrac{a+b}{2} \implies (a+b) = 2AM$ --- (i)

And,
$GM = \sqrt{ab}$
Also,
$HM = \dfrac{2ab}{a+b}$
Given,
$GM = HM + 1.6$  --- (ii)

$AM = GM + 2$ --- (iii)
Solving equation (ii), we get,

$or, \sqrt{ab} = \dfrac{2ab}{a+b} + 1.6$

$or, \sqrt{ab} = \dfrac{2ab}{2(AM)} + 1.6$

$or, \sqrt{ab} = \dfrac{2ab}{2(GM + 2)} + 1.6$

$or, \sqrt{ab} = \dfrac{ab}{\sqrt{ab}+2} +  1.6$

$or, \sqrt{ab}(\sqrt{ab} + 2) = ab + 1.6(\sqrt{ab} + 2)$

$or, ab + 2 \sqrt{ab} = ab + 1.6\sqrt{ab} + 3.2$

On solving, we get, $\sqrt{ab} = 8$ - (iv)

Put value of $\sqrt{ab} = 8$ in equation (iii), we get,
$or, AM = 8 + 2 \implies (a+b) =20$

$or, b = 20 -a $ --- (v)
Put b = 20 - a$ in equation (iv), we get,
$\sqrt{a(20 - a)} = 8$
$or, 20  a - a^2 = 64$
$or, a^2 - 20a + 64 = 0$
$or, (a - 16)(a -4) = 0$
Either, $(a-16) = 0 \implies a = 16$
So, b = 4
Or, $(a-4) =0 \implies a = 4$
So, b = 16
Hence,
the required numbers (a,b) are either (16,4) or (4,16).


8) The sum of three numbers in AP is 36. When the numbers are increased by 1, 4, 43 respectively, the resulting numbers are in GP. Find the numbers.

Solution:

Let the three numbers in AP be $(a-d), a, (a+d)$

We have,
Sum of three numbers is 36
$or, (a-d) + a + (a+d) = 36$
$or, 3a = 36$
$\therefore a = 12 $
Given,
$(a-d+1), (a+4), (a+d+43)$ are in GP

Using formula of geometric mean, we get,

$or, (a+4)^2 = (a-d+1)(a+d+43)$

Put a = 12 in each values of a

$or, (12+4)^2 = (12-d+1)(12+d+43)$

$or, 16^2 = (13-d)(55+d)$
$or, 256 = 715 - 42d - d^2$
$or, d^2 +42d - 459 = 0$
$or, (d+51)(d-9) = 0$
Either, $(d + 51) = 0 \implies d= -51$

Or, $(d-9) = 0 \implies d= 9$
When d = -51, we get,
$(a-d) = 63, a = 12, (a+d) = -39$

When d = 9, we get,
$(a-d) = 3, a = 12, (a+d) = 21$

Hence, the required numbers are either (63,12,-39) or (3,12,21).













About the Textbook:

Name: Basic Mathematics Grade XI
Author(s): D.R. Bajracharya | R.M. Shresththa | M.B. Singh | Y.R. Sthapit | B.C. Bajracharya
Publisher: Sukunda Pustak Bhawan (Bhotahity, Kathmandu)
Telephone: 5320379, 5353537
Price: 695 /- (2078 BS)