4 Marks Questions - Maths STD 10 Questions - Vidyadip

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Question 14 Marks

On the world environment day tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.

Answer

First term a = 1
Second term $\mathrm{t}_1=2$
Third term $\mathrm{t}_3=3$
Common difference $d=t_3-t_2=3-2=1$
We need to find total number of trees when $n=25$
Thus, By using sum of $n^{\text {th }}$ term of an A.P. we will find it's sum
$S_n=\frac{n}{2}[2 a+(n-1) d]$
Where, $n=$ no. of terms
$a=$ first term
$\mathrm{d}=$ common difference
$S_n=$ sum of $n$ terms
We need to find $\mathrm{S}_{25}$
Thus, on substituting the given value in formula we get,
$\Rightarrow S_{25}=\frac{25}{2}[2 \times 1+(25-1) \times 1] $
$ \Rightarrow S_{25}=\frac{25}{2}[2+24] $
$ \Rightarrow S_{25}=\frac{25}{2} \times 2 \times[1+12] $
$ \Rightarrow S_{25}=25 \times 13=325$

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Question 24 Marks

Kargil's temperature was recorded in a week from Monday to Saturday. All readings were in A.P. The sum of temperatures of Monday and Saturday was $5^{\circ} C$ more than sum of temperatures of Tuesday and Saturday. If temperature of Wednesday was $-30^{\circ}$ celsius then find the temperature on the other five days.

Answer

Let Monday be the first term i.e. $a=t_1$
Let Tuesday be the second term i.e $t _2$
Let Wednesday be the third term i.e $t _3$
Let Thursday be the fourth term i.e $t _4$
Let Friday be the fifth term i.e $t _5$
Let Saturday be the sixth term i.e $t _6$
$\text { Given: } t_1+t_6=5+\left(t_2+t_6\right)$
$\Rightarrow a=5+\left(t_2+t_6\right)-t_6$
$\Rightarrow a=5+t_2 \ldots \ldots(1)$
We know that,
Now, By using $n ^{\text {th }}$ term of an A.P. formula
$t_n=a+(n-1) d$
where $n=n o$. of terms
$a=\text { first term }$
$d=\text { common difference }$
$t_n=n^{\text {th }} \text { terms }$
Thus, $t _2= a +(2-1) d$
$\Rightarrow t_2=a+d$
Now substitute value of $t _2$ in (1) we get,
$\Rightarrow a=5+(a+d)$
$\Rightarrow d=a-5-a=-5$
Given: $t _3=-30^{\circ}$
$\text { Thus, } t_3=a+(3-1) \times(-5)$
$\Rightarrow-30=a+2 \times(-5)$
$\Rightarrow-30=a-10$
$\Rightarrow a=-30+10=-20^{\circ}$
Thus, Monday, $a=t_1=-20^{\circ}$
Using formula $t_{n+1}=t_n+d$
We can find the value of the other terms
Tuesday, $t _2= t _1+ d =-20-5=-25^{\circ}$
Wednesday, $t _3= t _2+ d =-25-5=-30^{\circ}$
Thursday, $t_4=t_3+d=-30-5=-35^{\circ}$
Friday, $t _5= t _4+ d =-35-5=40^{\circ}$
Saturday, $t_6=t_5+d=-40-5=-45^{\circ}$
Thus, we obtain an A.P.
$-20^{\circ},-25^{\circ},-30^{\circ},-35^{\circ},-40^{\circ},-45^{\circ}$

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Question 34 Marks

There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the 15th row and also find how many total seats are there in the auditorium?

Answer

Given: first term $\mathrm{a}=20$
Second term $t_1=22$
Third term $\mathrm{t}_2=24$
Common difference $d=t_2-t_1=24-22=2$
We need to find $t_{15}$ thus $n=15$
Now, By using $\mathrm{n}^{\text {th }}$ term of an A.P. formula
$t_n=a+(n-1) d$
where $n=$ no. of terms
$a=$ first term
$\mathrm{d}=$ common difference
$\mathrm{t}_{\mathrm{n}}=\mathrm{n}^{\text {th }} \text { terms }$
On substituting all value in $\mathrm{n}^{\text {th }}$ term of an A.P.
$\Rightarrow t_{15}=20+(15-1) \times 2$
$\Rightarrow t_{15}=20+14 \times 2$
$\Rightarrow t_{15}=20+28=48$
We have been given that, there are 27 rows in an auditorium
Thus, we need to find total seats in auditorium i.e. $\mathrm{S}_{27}$
Now, By using sum of $n^{\text {th }}$ term of an A.P. we will find it's sum
$S_n=\frac{n}{2}[2 a+(n-1) d]$
Where, $\mathrm{n}=$ no. of terms
$a=$ first term
$\mathrm{d}=$ common difference
$\mathrm{S}_{\mathrm{n}}=$ sum of $n$ terms
Thus, on substituting the given value in formula we get,
$\Rightarrow S_{27}=\frac{27}{2}[2 \times 20+(27-1) \times 2]$
$\Rightarrow S_{27}=\frac{27}{2} \times 2 \times[20+26]$
$\Rightarrow S_{27}=27 \times 46$
$\Rightarrow S_{27}=1242$

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Question 44 Marks

Sachin invested ina national saving certificate scheme. In the first year he invested ₹ 5000, in the second year ₹ 7000, in the third year ₹ 9000 and so on. Find the total amount that he invested in 12 years.

Answer

By given information we can form an A.P.
$5000,7000,9000, \ldots \ldots .$
Hence, the first term $a=5000$
Second term $\mathrm{t}_1=7000$
Third term $\mathrm{t}_2=9000$
Thus, common difference $d=t_2-t_1=9000-7000=2000$
Here, number of terms $\mathrm{n}=12$
We need to find $\mathrm{S}_{12}$
Now, By using sum of $\mathrm{n}^{\text {th }}$ term of an A.P. we will find it's sum
$S_n=\frac{n}{2}[2 a+(n-1) d]$
Where, $\mathrm{n}=$ no. of terms
$a=$ first term
$\mathrm{d}=$ common difference
$\mathrm{S}_{\mathrm{n}}=$ sum of n terms
Thus, on substituting the given value in formula we get,
$\Rightarrow S_{12}=\frac{12}{2}[2 \times 5000+(12-1) \times 2000]$
$\Rightarrow S_{12}=6 \times[10,000+11 \times 2000]$
$\Rightarrow S_{12}=6 \times[10,000+22,000]$
$\Rightarrow S_{12}=6 \times 32,000$
$\Rightarrow S_{12}=\text { Rs. } 192000$

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Question 54 Marks

A man borrows ₹ 8000 and agrees to repay with a total interest of ₹ 1360 in 12 monthly instalments. Each instalment being less than the preceding one by ₹ 40. Find the amount of the first and last instalment.

Answer

Given: A man borrows = Rs. 8000
Repay with total interest = Rs 1360
In 12 months, thus n = 12
Thus, $S_{12}$ = 8000 + 1360 = 9360
Each installment being less than preceding one
Thus, d = – 40
We need to find “a”
Now, By using sum of $n^{th}$ term of an A.P. we will find it’s sum
$S_n = \frac{n}{2} [2a+(n-1)]d$
Where, n = no. of terms
a = first term
d = common difference
$S _{ n }$ = sum of n terms
Thus, on substituting the given value in formula we get,
$\Rightarrow S_{12}=\frac{12}{2}[2 a+(12-1) \times(-40)]$
$\Rightarrow 9360=6[2 a-11 \times 40] $
$\Rightarrow \frac{9360}{6}=2 a-440 $
$\Rightarrow 1560=2 a-440 $
$\Rightarrow 1560+440=2 a $
$\Rightarrow 2 a=2000 $
$\Rightarrow a=\frac{2000}{2}=1000$
Thus, first installment $a=$ Rs. 1000
Now, By using sum of $n ^{\text {th }}$ term of an A.P. we will find it's sum
$S _{ n }=\frac{ n }{2}[\text { first term }+ \text { last term }]$
Where, $n=$ no. of terms
$S _{ n }=$ sum of n terms
Thus, on substituting the given value in formula we get,
Let $a=$ first term, $t_n=$ last term
$\Rightarrow S _{12}=\frac{12}{2}\left[ a + t _{ n }\right] $
$\Rightarrow 9360=6\left[1000+ t _{ n }\right] $
$\Rightarrow 1000+ t _{ n }=\frac{9360}{6}=1560 $
$\Rightarrow t _{ n }=1560-1000=560$
Thus, last installment $t _{ n }=560$

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Question 64 Marks

On $1^{st}$ Jan 2016, Sanika decides to save ₹ 10, ₹ 11 on second day, ₹ 12 on third day. If she decides to save like this, then on 31st Dec 2016 what would be her total saving?

Answer

By given information we can form an A.P.
$10,11,12,13, \ldots \ldots .$
Hence, the first term $a=10$
Second term $t_1=11$
Third term $\mathrm{t}_2=12$
Thus, common difference $d=t_2-t_1=12-11=1$
Here, number of terms from $1^{\text {st }}$ Jan 2016 to $31^{\text {st }}$ Dec 2016 is,
$\mathrm{n}=366$
We need to find $\mathrm{S}_{366}$
Now, By using sum of $\mathrm{n}^{\text {th }}$ term of an A.P. we will find it's sum
$S_n=\frac{n}{2}[2 a+(n-1) d]$
Where, $\mathrm{n}=$ no. of terms
$a=$ first term
$\mathrm{d}=$ common difference
$\mathrm{S}_{\mathrm{n}}=$ sum of n terms
Thus, on substituting the given value in formula we get,
$\Rightarrow S_{366}=\frac{366}{2}[2 \times 10+(366-1) \times 1]$
$\Rightarrow S_{366}=183[20+365]$
$\Rightarrow S_{366}=183 \times 385$
$\Rightarrow S_{366}=\operatorname{Rs} 70,455$

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Question 74 Marks

If the $9 \text {th}$ term of an A.P. is zero then show that the $29^{th}$ term is twice the $19^{th}$ term.

Answer

Now, By using $n ^{\text {th }}$ term of an A.P. formula
$t_n=a+(n-1) d$
where $n=n o$. of terms
$a=$ first term
$d =$ common difference
$t _{ n }= n ^{\text {th }}$ terms
Given: $t_9=0$
$\Rightarrow t_9=a+(9-1) d$
$\Rightarrow 0=a+8 d$
$\Rightarrow a=-8 d$
To Show: $t _{29}=2 \times t _{19}$
Now,
$\Rightarrow t_{29}=a+(29-1) d$
$\Rightarrow t_{29}=a+28 d$
$\left.\Rightarrow t_{29}=-8 d+28 d=20 d \text { (since, } a=-8 d\right)$
$\Rightarrow t_{29}=20 d$
$\Rightarrow t_{29}=2 \times 10 d \ldots . .(1)$
Also,
$\Rightarrow t_{19}=a+(19-1) d$
$\Rightarrow t_{19}=a+18 d$
$\Rightarrow t_{19}=-8 d+18 d=10 d(\text { since, } a=-8 d)$
$\Rightarrow t_{19}=10 d \ldots . . .(2)$
From eq. (1) and eq. (2) we get,
$t_{29}=2 \times t_{19}$

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Question 84 Marks

Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14.
(Assume the four consecutive terms in A.P. are a – d, a, a + d, a + 2d.)

Answer

Let the first term be a – d
the second term be a
the third term be a + d
the fourth term be a + 2 d
Given: sum of consecutive four term is 12
⇒ (a – d) + a + (a + d) + (a + 2d) = 12
⇒ 4 a + 2d = 12
⇒ 2(2 a + d) = 12
$\Rightarrow 2a +d = \frac{12}{2} = 6$
⇒ 2a + d = 6 …..(1)
Also, sum of third and fourth term is 14
⇒ (a + d) + (a + 2d) = 14
⇒ 2a + 3d = 14 ……(2)
Subtracting eq. (1) from eq. (2) we get,
⇒(2a + 3d) – (2a + d) = 14 – 6
⇒2a + 3d – 2a – d = 8
⇒ 2d = 8
$\Rightarrow d = \frac{8}{2} = 4$
⇒ d = 4
Substituting value of “d” in eq. (1) we get,
⇒ 2a + 4 = 6
⇒ 2a = 6 – 4 = 2
$\Rightarrow a = \frac{2}{2} = 1$
⇒ a = 1
Thus, a = 1 and d = 4
Hence, first term a – d = 1 – 4 = – 3
the second term a = 1
the third term a + d = 1 + 4 = 5
the fourth term a + 2 d = 1 + 2×4 = 1 + 8 = 9
Thus, the A.P. is – 3, 1, 5, 9

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Question 94 Marks

In an A.P. sum of three consecutive terms is 27 and their product is 504 find the terms? (Assume that three consecutive terms in A.P. are a – d, a, a + d.)

Answer

Let the first term be $a-d$ the second term be a the third term be $a+d$
Given: sum of consecutive three term is 27
$\Rightarrow(a-d)+a+(a+d)=27$
$\Rightarrow 3 a=27$
$\Rightarrow a=\frac{27}{3}=9$
Also, Given product of three consecutive term is 504
$\Rightarrow(a-d) \times a \times(a+d)=504$
$\Rightarrow(9-d) \times 9 \times(9+d)=504(\text { since, } a=9)$
$\Rightarrow(9-d) \times(9+d)=\frac{504}{9}=56$
$\Rightarrow 9^2-d^2=56\left(\text { since, }(a-b)(a+b)=a^2-b^2\right)$
$\Rightarrow 81-d^2=56$
$\Rightarrow d^2=81-56=25$
$\Rightarrow d=\sqrt{ } 25= \pm 5$
Case 1:
Thus, if $a=9$ and $d=5$
Then the three terms are,
First term $a-d=9-5=4$
Second term $a=9$
Third term $a+d=9+5=14$
Thus, the A.P. is $4,9,14$
Case 2:
Thus, if $a=9$ and $d=-5$
Then the three terms are,
First term $a-d=9-(-5)=9+5=14$
Second term $a=9$
Third term $a+d=9+(-5)=9-5=4$
Thus, the A.P. is $14,9,4$

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Question 104 Marks

Sum of first 55 terms in an A.P. is 3300, find its 28th term.

Answer

Given: $S_{55}=3300$ where $n=55$
Now, By using sum of $n^{\text {th }}$ term of an A.P. we will find it's sum
$S_n=\frac{n}{2}[2 a+(n-1)] d$
Where, $\mathrm{n}=$ no. of terms
$a=$ first term
$\mathrm{d}=$ common difference
$\mathrm{S}_{\mathrm{n}}=$ sum of $n$ terms
Thus, on substituting the given value in formula we get,
$\Rightarrow S_{55}=\frac{55}{2}[2 a+(55-1) d]$
$\Rightarrow 3300=\frac{55}{2}[2 a+54 d]$
$\Rightarrow 3300=\frac{55}{2} \times 2 \times[a+27 d]$
$\Rightarrow 3300=55 \times[a+27 d]$
$\Rightarrow \frac{3300}{55}=a+27 d$
$\Rightarrow a+27 \mathrm{~d}=60 \ldots \ldots(1)$
We need to find value of $28^{\text {th }}$ term i.e $\mathrm{t}_{28}$
Now, By using $\mathrm{n}^{\text {th }}$ term of an A.P. formula
$t_n=a+(n-1) d$
where $n=$ no. of terms
$\mathrm{a}=$ first term
$\mathrm{d}=$ common difference
$\mathrm{t}_{\mathrm{n}}=\mathrm{n}^{\text {th }} \text { terms }$
we can find value of $t_{28}$ by substituting all the value in formula we get,
$\Rightarrow t_{28}=a+(28-1) d$
$\Rightarrow t_{28}=a+27 d$
From eq. (1) we get,
$\Rightarrow t_{28}=a+27 d=60$
$\Rightarrow t_{28}=60$

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Question 114 Marks

In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.

Answer

Given: $\mathrm{t}_{19}=52$ and $\mathrm{t}_{38}=128$
To find: value of "a" and "d"
Using $n^{\text {th }}$ term of an A.P. formula
$t_n=a+(n-1) d$
$\text { where } n=\text { no. of terms }$
$a=\text { first term }$
$d=\text { common difference }$
$t_n=n^{\text {th }} \text { terms }$
we will find value of "a" and "d"
$\text { Let, } \mathrm{t}_{19}=a+(19-1) \mathrm{d}$
$\Rightarrow 52=a+18 d \ldots . .(1)$
$t_{38}=a+(38-1) d$
$\Rightarrow 128=a+37 d \ldots . .(2)$
Subtracting eq. (1) from eq. (2), we get,
$\Rightarrow 128-52=(a-a)+(37 d-18 d)$
$\Rightarrow 76=19 d$
$\Rightarrow d=\frac{76}{19}=4$
Substitute value of "d" in eq. (1) to get value of "a"
$\Rightarrow 52=a+18 \times 4$
$\Rightarrow 52=a+72$
$\Rightarrow a=52-72=-20$
Now, to find value of $S_{56}$ we will using formula of sum of $n$ terms
$S_n=\frac{n}{2}[2 a+(n-1)] d$
Where, $\mathrm{n}=$ no. of terms
$a=$ first term
$\mathrm{d}=$ common difference
$\mathrm{S}_{\mathrm{n}}=$ sum of $n$ terms
Thus, Substituting given value in formula we can find the value of $S_n$
$\Rightarrow S_{56}=\frac{56}{2}[2 \times(-20)+(56-1) \times 4]$
$\Rightarrow S_{56}=28 \times[-40+55 \times 4]$
$\Rightarrow S_{56}=28 \times[-40+220]$
$\Rightarrow S_{56}=28 \times 180=5040$
Thus, $\mathrm{S}_{56}=5040$

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Question 124 Marks

Find the sum of all even numbers from 1 to 350.

Answer

List of even natural number between 1 to 350 is $2,4,6, \ldots . . . .348$
Where first term $a=2$
Second term $\mathrm{t}_1=4$
Third term $\mathrm{t}_2=6$
Thus, common difference $d=t_2-t_1=6-4=2$
$t_n=348$ (As we have to find the sum of even numbers between 1 and 350 therefore excluding 350 )
Now, By using $n^{\text {th }}$ term of an A.P. formula
$t_n=a+(n-1) d$
where $\mathrm{n}=$ no. of terms
$a=$ first term
$\mathrm{d}=$ common difference
$\mathrm{t}_{\mathrm{n}}=\mathrm{n}^{\text {th }} \text { terms }$
we can find value of " n " by substituting all the value in formula we get,
$\Rightarrow 348=2+(\mathrm{n}-1) \times 2$
$\Rightarrow 348-2=2(\mathrm{n}-1)$
$\Rightarrow 346=2(\mathrm{n}-1)$
$\Rightarrow n-1=\frac{346}{2}=173$
$\Rightarrow \mathrm{n}=173+1=174$
Now, By using sum of $n^{\text {th }}$ term of an A.P. we will find it's sum
$S_n=\frac{n}{2}[2 a+(n-1)] d$
Where, $\mathrm{n}=$ no. of terms
$a=$ first term
$\mathrm{d}=$ common difference
$S_n=$ sum of $n$ terms
Thus, Substituting given value in formula we can find the value of $S_n$
$\Rightarrow S_{174}=\frac{174}{2}[2 \times 2+(174-1) \times 2]$
$\Rightarrow S_{174}=\frac{174}{2}[4+173 \times 2]$
$\Rightarrow S_{174}=\frac{174}{2}[4+346]$
$\Rightarrow S_{174}=\frac{174}{2} \times 350=174 \times 175=30,450$
Thus, $\mathrm{S}_{174}=30,450$

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Question 134 Marks

Find the sum of first 123 even natural numbers.

Answer

List of first 123 even natural number is
2,4,6,......
Where first term $a=2$
Second term $\mathrm{t}_1=4$
Third term $\mathrm{t}_2=6$
Thus, common difference $d=t_2-t_1=6-4=2$
$n=123$
By using sum of $\mathrm{n}^{\text {th }}$ term of an A.P. is
$S_n=\frac{n}{2}[2 a+(n-1) d]$
Where, $\mathrm{n}=$ no. of terms
$a=$ first term
$\mathrm{d}=$ common difference
$S_n=$ sum of $n$ terms
Thus, Substituting given value in formula we can find the value of $S_n$
$\Rightarrow S_n=\frac{123}{2}[2 \times 2+(123-1) \times 2] $
$ \Rightarrow S_n=\frac{123}{2}[4+122 \times 2]$
$ \Rightarrow S_n=\frac{123}{2}[4+244] $
$ \Rightarrow S_n=\frac{123}{2} \times 248=123 \times 122=15252$
Thus, $S_n$ = 15252

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Question 144 Marks

If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero. (p ≠ q)

Answer

We know that, sum of $\mathrm{n}^{\text {th }}$ term of an A.P. we will find it's sum
$S_n=\frac{n}{2}[2 a+(n-1) d]$
Where, $n=$ no. of terms
$a=$ first term
$\mathrm{d}=$ common difference
$\mathrm{S}_{\mathrm{n}}=$ sum of n terms
Now, Sum of $p$ terms is
$S_p=\frac{p}{2}[2 a+(p-1) d]$
And, Sum of q terms is
$S_q=\frac{q}{2}[2 a+(q-1) d]$
Given: $S_p=S_q$
$\Rightarrow \frac{p}{2}[2 a+(p-1) d]=\frac{q}{2}[2 a+(q-1) d]$
Multiply by 2 on both sides, we get,
$\Rightarrow p[2 a+(p-1) d]=q[2 a+(q-1) d]$
$\Rightarrow 2 a p+p(p-1) d=2 a q+q(q-1) d$
$\Rightarrow 2 a p-2 a q+p(p-1) d-q(q-1) d=0$
$\Rightarrow 2 a(p-q)+d\left[p^2-p-q^2+q\right]=0$
$\Rightarrow 2 a(p-q)+d\left[\left(p^2-q^2\right)-p+q\right]=0$
$\Rightarrow 2 a(p-q)+d[(p-q)(p+q)-(p-q)]=0$
$\left(\text { since, }(a-b)(a+b)=a^2-b^2\right)$
$\Rightarrow 2 a(p-q)+d(p-q)[p+q-1]=0$
$\Rightarrow(p-q)[2 a+d(p+q-1)]=0$
Since, $p \neq q$
$\therefore p-q \neq 0$
$\Rightarrow 2 a+d(p+q-1)=0$
Multiply both side by $\frac{p+q}{2}$
$\Rightarrow \frac{p+q}{2}[2 a+d(p+q-1)]=0$
$\Rightarrow S_{p+q}=0$
Hence proved

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Question 154 Marks

If first term of an A.P. is a, second term is b and last term is c, then show that sum of all terms is $\frac{(a+c)(b+c-2 a)}{2(b-a)}$

Answer

Given first term = a
Second term $=b$
Last term $=\mathrm{c}$
Common difference $d=$ second term - first term $=b-a$
We will first find the number of terms
We use $\mathrm{n}^{\text {th }}$ term of an A.P. formula
$t_n=a+(n-1) d$
where $n=$ no. of terms
$a=$ first term
$\mathrm{d}=$ common difference
$\mathrm{t}_{\mathrm{n}}=\mathrm{n}^{\mathrm{th}} \text { terms }$
Thus, on substituting all values we get,
$\Rightarrow c=a+(n-1)(b-a)$
$\Rightarrow c=a+(b-a) n+a-b$
$\Rightarrow c=2 a-b+(b-a) n$
$\Rightarrow(b-a) n=c+b-2 a$
$\Rightarrow n =\frac{ c + b -2 a }{ b - a }$
Using Sum of n terms of an A.P. formula
$S _{ n }=\frac{ n }{2}[\text { first term }+ \text { last term }]$
where $n=$ no. of terms
$S_n=\text { sum of } n \text { terms} $
On substituting all the values we get,
$\Rightarrow S_n=\frac{c+b-2 a}{2(b-a)}[a+c] $
$\Rightarrow S_n=\frac{(a+c)(c+b-2 a)}{2(b-a)}$
Hence, proved

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Question 164 Marks

There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.

Answer

Let first term = a
Common difference $=\mathrm{d}$
Since, A.P. consist of 37 terms, therefor the middle most term is $\frac{37+1}{2}=\frac{38}{2}=19^{\text {th }}$ thrm,
Thus, three middle most term are $\mathrm{t}_{18}=18^{\text {th }}$ term, $\mathrm{t}_{19}=19^{\text {th }}$ term, $\mathrm{t}_{20}=20^{\text {th }}$ term
We use $\mathrm{n}^{\text {th }}$ term of an A.P. formula
$t_n=a+(n-1) d$
where $n=$ no. of terms
$a=\text { first term }$
$d=\text { common difference }$
$t_n=n^{\text {th }} \text { terms }$
Thus, on substituting all values we get,
$\text { Given, } \mathrm{t}_{18}+\mathrm{t}_{19}+\mathrm{t}_{20}=225$
$\Rightarrow[\mathrm{a}+(18-1) \mathrm{d}]+[\mathrm{a}+(19-1) \mathrm{d}]+[\mathrm{a}+(20-1) \mathrm{d}]=225$
$\Rightarrow[\mathrm{a}+17 \mathrm{~d}]+[\mathrm{a}+18 \mathrm{~d}]+[\mathrm{a}+19 \mathrm{~d}]=225$
$\Rightarrow 3 \mathrm{a}+54 \mathrm{~d}=225$
Dividing by 3
$\Rightarrow a+18 d=75 \ldots \ldots(1)$
Given, sum of last three term is 429
$\Rightarrow t_{35}+t_{36}+t_{37}=429$
$\Rightarrow[a+(35-1) d]+[a+(36-1) d]+[a+(37-1) d]=429$
$\Rightarrow[a+34 d]+[a+35 d]+[a+36 d]=429$
$\Rightarrow 3 a+105 d=429$
Dividing by 3
$\Rightarrow a+35 d=143 \ldots \ldots(2)$
Subtracting eq. (1) from eq. (2) we get,
$\Rightarrow[\mathrm{a}+35 \mathrm{~d}]-[\mathrm{a}+18 \mathrm{~d}]=143-75$
$\Rightarrow 17 \mathrm{~d}=68$
$\Rightarrow d=\frac{68}{17}=4$
Substituting value of 'd' in eq. (1) we get,
$\Rightarrow a+18 \times 4=75$
$\Rightarrow a+72=75$
$\Rightarrow a=75-72=3$
$\Rightarrow a=t_1=3$
We know that, $t_{n+1}=t_n+d$
$\mathrm{t}_2=\mathrm{t}_1+\mathrm{d}=3+4=7$
$\mathrm{t}_3=\mathrm{t}_2+\mathrm{d}=7+4=11$
$\mathrm{t}_4=\mathrm{t}_3+\mathrm{d}=11+4=15$
$\mathrm{t}_{37}=3+(37-1) \times 4$
$\mathrm{t}_{37}=3+36 \times 4$
$\mathrm{t}_{37}=3+144=147$
Thus, the A.P. is $3,7,11, \ldots, 147$

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Question 174 Marks

In an A.P. the first term is – 5 and last term is 45. If sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?

Answer

Given, first term $a=-5$
Last term $t_n=45$
Sum of $n$ terms $S_n=120$
To find no of terms " n "
Using Sum of $n$ terms of an A.P. formula $S_n=\frac{n}{2}[$ first term + last term $]$
where $n=$ no. of terms
$S_n=\text { sum of } \mathrm{n} \text { terms }$
Now, on substituting given value in formula we get,
$\Rightarrow 120=\frac{n}{2}[-5+45]$
$\Rightarrow 120=\frac{n}{2} \times 40$
$\Rightarrow 120=20 n$
$\Rightarrow n=\frac{120}{20}=6$
To find the common difference ' $d$ '
We use $n^{\text {th }}$ term of an A.P. formula
$t_n=a+(n-1) d$
where $n=$ no. of terms
$a=$ first term
$d=$ common difference
$\mathrm{t}_{\mathrm{n}}=\mathrm{n}^{\text {th }} \text { terms }$
Thus, on substituting all values we get,
$\Rightarrow \mathrm{t}_6=-5+(6-1) \mathrm{d}$
$\Rightarrow 45=-5+5 \mathrm{~d}$
$\Rightarrow 5 \mathrm{~d}=45+5=50$
$\Rightarrow d=\frac{50}{5}=10$
Thus, common difference is 10

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Question 184 Marks

Two A.P.’s are given $9, 7, 5, . . .$ and $24, 21, 18, . . . .$ If nth term of both the progressions are equal then find the value of n and nth term.

Answer

Given A.P. is $9,7,5 \ldots$...
Whose first tern a $=9$
Second term $t _1=7$
Third term $t _3=5$
Common difference $d=t_3-t_2=5-7=-2$
Another A.P. is $24,21,18, \ldots$
Whose first tern a $=24$
Second term $t _1=21$
Third term $t _3=18$
Common difference $d=t_3-t_2=18-21=-3$
We have been given, $n ^{\text {th }}$ term of both the A.P. is same thus, by using $n ^{\text {th }}$ term of an A.P. formula
$t_n=a+(n-1) d$
where $n =$ no. of terms
$a=$ first term
$d =$ common difference
$t _{ n }= n ^{\text {th }}$ terms
Hence, by given condition we get,
$\Rightarrow 9+(n-1) \times(-2)=24+(n-1) \times(-3)$
$\Rightarrow 9-2 n+2=24-3 n+3$
$\Rightarrow 11-2 n=27-3 n$
$\Rightarrow 3 n-2 n=27-11$
$\Rightarrow n=16$
Thus, value of $n ^{\text {th }}$ term where $a=9, d=-2, n=16$ is
$\Rightarrow t_n=9+(16-1) \times(-2)$
$\Rightarrow t_n=9-15 \times 2$
$\Rightarrow t_n=9-30=-21$

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Question 194 Marks

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Question 204 Marks

Anvar saves some amount every month.In first three months he saves ₹ 200, ₹ 250 and ₹ 300 respectively. In which month will he save ₹ 1000?

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Question 214 Marks

Ajay sharma repays the borrowed amount of ₹ 3,25,000 by paying ₹ 30500 in the first month and then decreases the payment by ₹ 1500 every month. How long will it take to clear his amount?

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Question 224 Marks

A mixer manufacturing company manufactured $600$ mixers in $3^{rd}$^ year and in $7^{th}$^ year they manufactured 700 mixers. If every year there is same growth in the production of mixers then find
(i) Production in the first year
(ii) Production in $10^{th}$ year
(iii) Total production in first seven years.

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Question 234 Marks

The $10^{\text {th }}$ term and the $18^{\text {th }}$ term of an A.P. are $25$ and $41$ respectively then find $38^{\text {th }}$ term of that A.P., similarly if $n^{\text {th }}$ term is $99$ . Find the value of $n$.

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Question 244 Marks

Find the middle term of sequence formed by all three digit numbers which leave a remainder $3$ when divided by $4.$ Also find sum of all numbers on both sides of the middle term.

Answer

The sequence formed by the given numbers is $103, 107, 111, 115, .... , 999.$
This is an $A.P.$ with $a=103, d=107-103=4$ and
$t_{ n }=999$
$t_n=a+(n-1) d$
$\therefore 999=103+(n-1) 4$
$\therefore 896=4 n-4$
$\therefore 4 n=900$
$\therefore n=\frac{900}{4}$
$\therefore n=225$
$ \therefore \text { Middle term } =\left(\frac{n+1}{2}\right)^{ th } \text { term }$
$ =\left(\frac{225+1}{2}\right)^{ th } \text { term }$
$ =\frac{226}{2}$
$ =113^{ th } \text { term. }$
$t_n=a+(n-1) d$
$\therefore t_{113}=103+(113-1) 4$
$\therefore \quad=103+112 \times 4$
$=103+448$
$t_{113}=551$
Now, $t_{112}=551-4=547$
So, we have to find $S_{112}$ and $\left(S_{225}-S_{113}\right)$
$S _n=\frac{n}{2}\left[t_1+t_{ n }\right]$
$\therefore S _{112}=\frac{112}{2}[103+547]$
$\therefore \quad=\frac{112}{2} \times 650$
$\therefore S _{112}=36400$
Similarly,
$ S _{225}- S _{113} =\frac{225}{2}(103+999)-\frac{113}{2}(103+551)$
$ =(225 \times 551)-(113 \times 327)$
$ =123975-36951$
$ =87024$
$\therefore$ Sum of all members on $\text{LHS}$ of middle term is $36400$ and sum of all numbers on $\text{RHS}$ of the middle term is $87024 .$

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Question 254 Marks

Jinal saves $₹ 1600$ during the first year, $₹ 2100$ in the second year, $₹ 2600$ in the third year, If she continues her saving in this pattern, in how many years will she save $₹ 38,500 ?$

Answer

Pattern of her saving is $₹ 1600, ₹ 2100, ₹ 2600, ....$
It is an $A.P$. with $a=1600, d=2100-1600=500$
Also, $S_n=38500$
$S_n=\frac{n}{2}[2 a+(n-1) d]$
$\therefore 38500=\frac{n}{2}[2 \times 1600+(n-1) 500]$
$\therefore 77000=n[3200+500 n-500]$
$\therefore 77000=n[2700+500 n]$
$\therefore 77000=2700 n+500 n^2$
$\therefore 500 n^2+2700 n-77000=0$
$\therefore 5 n^2+27 n-770=0$ ...[Dividing both sides by 100]
$\therefore 5 n^2+77 n-50 n-770=0$
$\therefore n(5 n+77)-10(5 n+77)=0$
$\therefore \quad(n-10)(5 n+77)=0$
$\therefore n-10=0 \text { or } 5 n+77=0$
$\therefore n=10 \text { or } n=\frac{-77}{5}$
Now $n=\frac{-77}{5}$ (Not possible as $n$ is natural number)
$\therefore n=10$
$\therefore$ Jinal will save $₹ 38500$ in $10$ years.

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Question 264 Marks

If $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of an $A. P$. are $l, m, n$ respectively, show that
$(q-r) l+(r-p) m+(p-q) n=0$

Answer

Given : $t_p=l, t_q=m, t_r=n$
Let the $1^{s t}$ term of given $A.P$. be $a$ and $d$ be the common difference.
$ t_n =a+(n-1) d$
$\therefore t_p =a+(p-1) d$
$\therefore l=a+(p-1) d ...(i)$
$\therefore t_q=a+(q-1) d$
$\therefore m=a+(q-1) d ...(ii)$
$\therefore t_r=a+(r-1) d$
$\therefore n=a+(r-1) d ...(iii)$
Subtracting $(ii)$ from $(i),$ we get

$\therefore l-m=(p-1-q+1) d$
$\therefore l-m=(p-q) d$
$\therefore \frac{l-m}{d}=(p-q) ...(iv)$
Subtracting $(iii)$ from $(ii),$ we get

$\therefore m-n=(q-1-r+1) d$
$\therefore m-n=(q-r) d$
$\therefore \frac{m-n}{d}=(q-r) ...(v)$
Subtracting $(i)$ from $(iii),$ we get

$\therefore n-l=(r-1-p+1) d$
$\therefore n-l=(r-p) d$
$\therefore \frac{n-l}{d}=(r-p) ...(vi)$
Now, $(q-r) l+(r-p) m+(p-q) n$
$=\left(\frac{m-n}{d}\right) l+\left(\frac{n-l}{d}\right) m+\left(\frac{l-m}{d}\right) n ...[$From $(iv), (v), (vi)]$
$=\frac{1}{d}[l m-\ln +m n-\operatorname{lm}+\ln -m n]$
$= 0$
Hence proved.

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Question 274 Marks

How many terms of the A.P. 16, 14, 12, ..... Are needed to given the sum 60? Explain why we get two answers.

Answer

For the A.P. 16, 14, 12, .....
a = 16, d = 14 – 16 = –2
Let $S_n=60$
$S_n=\frac{n}{2}[2 a+(n-1) d]$
$\therefore 60=\frac{n}{2}[2 \times 16+(n-1)(-2)]$
$\begin{array}{ll}\therefore 120=n[32-2 n+2] \\ \therefore 120=n[34-2 n] \\ \therefore 120=34 n-2 n^2 \\ \therefore 2 n^2-34 n+120=0\end{array}$
$\therefore \quad n^2-17 n+60=0$ ...[Dividing both sides by 2]
$\begin{array}{lr}\therefore n^2-12 n-5 n+60=0 \\ \therefore n(n-12)-5(n-12)=0 \\ \therefore (n-12)(n-5)=0 \\ \therefore n-12=0 \text { or } n-5=0 \\ \therefore n=12 \text { or } n=5\end{array}$
$\therefore$ The number of terms required to give a sum of 60 are 12 or 5 .
The reason for getting two answers is the common difference is –2 and the progression is towards the negative side.
The A. P. is as follows:
16, 14, 12, 10, 8, 6, 4, 2, 0, –2, –4, –6.....
$S_5=16+14+12+10+8=60$
$S_{12}=16+14+12+10+8+6+4+2+0-2$ $-4-6=60$

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Question 284 Marks

If sum of $m$ terms is $n$ and sum of $n$ terms is $m$, then show that the sum of $(m+n)$ terms is $-(m+n)$.

Answer

Given: $S_m=n, S_n=m$
To prove: $S_{m+n}=-(m+n)$
Proof:
$S_m=\frac{m}{2}[2 a+(m-1) d]$
$\therefore n=\frac{m}{2}[2 a+(m-1) d]...(i)$
$S_n=\frac{n}{2}[2 a+(n-1) d]$
$m=\frac{n}{2}[2 a+(n-1) d]...(ii)$
Subtracting $(ii)$ from $(i)$
$\therefore n-m=\frac{m}{2}[2 a+(m-1) d]-\frac{n}{2}[2 a+(n-1) d]$
$\therefore 2(n-m)=m[2 a+(m-1) d]-n[2 a+(n-1) d]$
$\therefore 2(n-m)=m[2 a+m d-d]-n[2 a+n d-d]$
$\therefore 2(n-m)=2 a m+m^2 d-m d-2 a n-n^2 d+n d$
$\therefore 2(n-m)=2 a m-2 a n+m^2 d-n^2 d-m d+n d$
$\therefore 2(n-m)=2 a(m-n)+d\left(m^2-n^2\right)-d(m-n)$
$\therefore \quad-2(m-n)=2 a(m-n)+d(m+n)(m-n)-d(m-n)$
Dividing throughout by $m – n,$
$\therefore \quad-2=2 a+d(m+n)-d$
$\therefore \quad-2=2 a+d(m+n-1) ...(iii)$
Now, $S_{m+n}=\frac{m+n}{2}[2 a+d(m+n-1)]$
$\therefore S_{m+n}=\frac{m+n}{2}[-2] ....[$From $(iii)]$
$\therefore S_{m+n}=m+n(-1)$
$\therefore S_{m+n}=-(m+n)$

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Question 294 Marks

$200$ logs of wood are stacked in the following manner: $20$ logs in the bottom row, $19$ in the next row, $18$ in the row next to it and so on. In how many rows $200$ logs are placed and how many logs are there in the top row?

Answer

The arrangement of logs $20,19,18, \ldots$. forms an $A.P.$ with $a=20, d=-1$
Let $200$ logs be arranged in n rows.
$S_n=200$
$S_n=\frac{n}{2}[2 a+(n-1) d]$
$\therefore 200=\frac{n}{2}[2 \times 20+(n-1)(-1)]$
$\therefore 400=n[40-n+1]$
$\therefore 400=n[41-n]$
$\therefore 400=41 n-n^2$
$\therefore n^2-41 n+400=0$
$\therefore n^2-25 n-16 n+400=0$
$\therefore n(n-25)-16(n-25)=0$
$\therefore \quad(n-25)(n-16)=0$
$\therefore n-25=0$ or $n-16=0$
$\therefore n=25$ or $n=16$
If $n=25$
$t_n=a+(n-1) d$
$\therefore t_{25}=20+(25-1)(-1)$
$=20+24(-1)$
$=20-24$
$t_{25}=-4$
No. of logs in the $25^{ th }$ row cannot be negative
$\therefore n \neq 25 \therefore n=16$
$t_n=a+(n-1) d$
$ \therefore t_{16} =20+(16-1)(-1)$
$=20+15(-1)$
$=20-15$
$t_{16} =5$
$\therefore 200$ logs are placed in $16$ rows and there are $5$ logs in the top row.

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Question 304 Marks

A farmer borrows ₹ 1000 and agrees to repay with a total interest of ₹ 140 in 12 installments, each installment being less than the preceding installment by ₹ 10 . What should be his first installment?

Answer

₹ 150

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Question 314 Marks

Mr. Shah borrows ₹ 4000 and agrees to repay with a total interest of ₹ 500 in 10 installments, each installment being less than the preceding installment by ₹ 10. What should be the first and the last installment?

Answer

₹ 495 , ₹ 405

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Question 324 Marks

Neeta saves in a 'Mahila Bachat Gat' ₹ 2 on the first day, ₹ 4 on the second day, ₹ 6 on the third day and so on. What will be her saving in the month of February 2010 ?

Answer

₹ 812

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Question 334 Marks

In winter, the temperature at a hill station from Monday to Friday is in A.P., The sum of the temperatures of Monday, Tuesday and Wednesday is zero and the sum of the temperature of Thursday and Friday is 15. Find the temperature of each of the five days.

Answer

$-3,0,3,6,9$

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Question 344 Marks

A meeting hall has 20 seats in the first row, 20 seats in the second, 28 seats in the third row and so on and has in all 30 rows. How many seats are there in the meeting hall?

Answer

2340

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Question 354 Marks

In an A.P. the first term is – 5 and last term is 45. If sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?

Answer

Given, first term a = – 5
Last term t_n = 45
Sum of n terms S_n = 120
To find no of terms “n”
Using Sum of n terms of an A.P. formula
$S _{ n }=\frac{ n }{2}[\text { first term }+ \text { last term }]$
where $n=$ no. of terms
$S_n=\text { sum of } n \text { terms }$
Now, on substituting given value in formula we get
$\begin{array}{l}
\Rightarrow 120=\frac{n}{2}[-5+45] \\
\Rightarrow 120=\frac{n}{2} \times 40 \\
\Rightarrow 120=20 n \\
\Rightarrow n=\frac{120}{20}=6
\end{array}$
To find the common difference ‘d’
We use n^th term of an A.P. formula
t_n = a + (n – 1)d
where n = no. of terms
a = first term
d = common difference
t_n = n^th terms
Thus, on substituting all values we get,
⇒ t₆ = – 5 + (6 – 1)d
⇒ 45 = – 5 + 5d
⇒ 5d = 45 + 5 = 50
$\Rightarrow d=\frac{50}{5}=10$
Thus, common difference is 10

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Question 364 Marks

Two A.P.’s are given 9, 7, 5, . . . and 24, 21, 18, . . . . If nth term of both the progressions are equal then find the value of n and nth term.

Answer

Given A.P. is 9, 7, 5….Whose first tern a = 9
Second term t₁ = 7
Third term t₃ = 5
Common difference d = t₃ – t₂ = 5 – 7 = – 2
Another A.P. is 24, 21, 18, . . .
Whose first tern a = 24
Second term t₁ = 21
Third term t₃ = 18
Common difference d = t₃ – t₂ = 18 – 21 = – 3
We have been given, n^th term of both the A.P. is same
thus, by using n^th term of an A.P. formula
t_n = a + (n – 1)d
where n = no. of terms
a = first term
d = common difference
t_n = n^th terms
Hence, by given condition we get,
⇒ 9 + (n – 1) × ( – 2) = 24 + (n – 1) × ( – 3)
⇒ 9 – 2n + 2 = 24 – 3n + 3
⇒ 11 – 2n = 27 – 3n
⇒ 3n – 2n = 27 – 11
⇒ n = 16
Thus, value of n^th term where a = 9, d = – 2, n = 16 is
⇒ t_n = 9 + (16 – 1) × ( – 2)
⇒ t_n = 9 – 15 × 2
⇒ t_n = 9 – 30 = – 21

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Question 374 Marks

If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero. (p ≠ q)

Answer

We know that, sum of $n^{\text {th }}$ term of an A.P. we will find it's sum
$S _{ n }=\frac{ n }{2}[2 a +( n -1) d ]$
Where, $n =$ no. of terms
$a=\text { first term }$
$d =$ common difference
$S _{ n }=\text { sum of } n \text { terms }$
Now, Sum of $p$ terms is
$S _{ p }=\frac{ p }{2}[2 a +( p -1) d ]$
And, Sum of $q$ terms is
$S _{ q }=\frac{ q }{2}[2 a +( q -1) d ]$
Given: $S _{ p }= S _{ q }$
$\Rightarrow \frac{p}{2}[2 a+(p-1) d]=\frac{q}{2}[2 a+(q-1) d]$
Multiply by 2 on both sides, we get,
$\begin{array}{l}
\Rightarrow p[2 a+(p-1) d]=q[2 a+(q-1) d] \\
\Rightarrow 2 a p+p(p-1) d=2 a q+q(q-1) d \\
\Rightarrow 2 a p-2 a q+p(p-1) d-q(q-1) d=0 \\
\Rightarrow 2 a(p-q)+d\left[p^2-p-q^2+q\right]=0 \\
\Rightarrow 2 a(p-q)+d\left[\left(p^2-q^2\right)-p+q\right]=0 \\
\Rightarrow 2 a(p-q)+d[(p-q)(p+q)-(p-q)]=0 \\
\left(\operatorname{since},(a-b)(a+b)=a^2-b^2\right) \\
\Rightarrow 2 a(p-q)+d(p-q)[p+q-1]=0 \\
\Rightarrow(p-q)[2 a+d(p+q-1)]=0
\end{array}$
$\begin{array}{l}
\therefore p-q \neq 0 \\
\Rightarrow 2 a+d(p+q-1)=0
\end{array}$
Multiply both side by $\frac{p+q}{2}$
$\begin{array}{l}
\Rightarrow \frac{p+q}{2}[2 a+d(p+q-1)]=0 \\
\Rightarrow S_{p+q}=0
\end{array}$
Hence proved

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Question 384 Marks

If first term of an A.P. is a, second term is b and last term is c, then show that sum of all terms is $\frac{(a+c)(b+c-2 a)}{2(b-a)}$.

Answer

Given first term = a
Second term = b
Last term = c
Common difference d = second term – first term = b – a
We will first find the number of terms
We use n^th term of an A.P. formula
t_n = a + (n – 1)d
where n = no. of terms
a = first term
d = common difference
t_n = n^th terms
Thus, on substituting all values we get,
⇒ c = a + (n – 1)(b – a)
⇒ c = a + (b – a)n + a – b
⇒ c = 2a – b + (b – a)n
⇒ (b – a)n = c + b – 2a
$\Rightarrow n =\frac{ c + b -2 a }{ b - a }$
Using Sum of $n$ terms of an A.P. formula
$S _{ n }=\frac{ n }{2} \text { [ first term + last term] }$
where $n =$ no. of terms
$S_n=\text { sum of } n \text { terms }$
On substituting all the values we get,
$\begin{array}{l}
\Rightarrow S _{ n }=\frac{ c + b -2 a }{2( b - a )}[ a + c ] \\
\Rightarrow S _{ n }=\frac{( a + c )( c + b -2 a )}{2( b - a )}
\end{array}$
Hence, proved

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Question 394 Marks

There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.

Answer

Let first term = a
Common difference = d
Since, A.P. consist of 37 terms, therefor the middle most term is
$\frac{37+1}{2}=\frac{38}{2}=19$ th term
Thus, three middle most term are t₁₈ = 18^th term, t₁₉ = 19^th term,
t₂₀ = 20^th term
We use n^th term of an A.P. formula
t_n = a + (n – 1)d
where n = no. of terms
a = first term
d = common difference
t_n = n^th terms
Thus, on substituting all values we get,
Given, t₁₈ + t₁₉ + t₂₀ = 225
⇒ [a + (18 – 1)d] + [a + (19 – 1)d] + [a + (20 – 1)d] = 225
⇒ [a + 17d] + [a + 18d] + [a + 19d] = 225
⇒ 3a + 54d = 225
Dividing by 3
⇒ a + 18d = 75 …….(1)
Given, sum of last three term is 429
⇒ t₃₅ + t₃₆ + t₃₇ = 429
⇒ [a + (35 – 1)d] + [a + (36 – 1)d] + [a + (37 – 1)d] = 429
⇒ [a + 34d] + [a + 35d] + [a + 36d] = 429
⇒ 3a + 105d = 429
Dividing by 3
⇒ a + 35d = 143 …….(2)
Subtracting eq. (1) from eq. (2) we get,
⇒ [a + 35d] – [a + 18d] = 143 – 75
⇒ 17d = 68
$\Rightarrow d=\frac{68}{17}=4$
Substituting value of ‘d’ in eq. (1) we get,
⇒ a + 18 × 4 = 75
⇒ a + 72 = 75
⇒ a = 75 – 72 = 3
⇒ a = t₁ = 3
We know that, t_{n + 1} = t_n + d
t₂ = t₁ + d = 3 + 4 = 7
t₃ = t₂ + d = 7 + 4 = 11
t₄ = t₃ + d = 11 + 4 = 15
t₃₇ = 3 + (37 – 1) × 4
t₃₇ = 3 + 36 × 4
t₃₇ = 3 + 144 = 147
Thus, the A.P. is 3, 7, 11, . . . . ., 147

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Question 404 Marks

Kargil’s temperature was recorded in a week from Monday to Saturday. All readings were in A.P. The sum of temperatures of Monday and Saturday was 5° C more than sum of temperatures of Tuesday and Saturday. If temperature of Wednesday was – 30° celsius then find the temperature on the other five days.

Answer

Let Monday be the first term i.e. a = t₁Let Tuesday be the second term i.e t₂
Let Wednesday be the third term i.e t₃
Let Thursday be the fourth term i.e t₄
Let Friday be the fifth term i.e t₅
Let Saturday be the sixth term i.e t₆
Given: t₁ + t₆ = 5 + (t₂ + t₆)
⇒ a = 5 + (t₂ + t₆) – t₆
⇒ a = 5 + t₂ …..(1)
We know that,
Now, By using n^th term of an A.P. formula
t_n = a + (n – 1)d
where n = no. of terms
a = first term
d = common difference
t_n = n^th terms
Thus, t₂ = a + (2 – 1) d
⇒ t₂ = a + d
Now substitute value of t₂ in (1) we get,
⇒ a = 5 + (a + d)
⇒ d = a – 5 – a = – 5
Given: t₃ = – 30°
Thus, t₃ = a + (3 – 1) × ( – 5)
⇒ – 30 = a + 2 × ( – 5)
⇒ – 30 = a – 10
⇒ a = – 30 + 10 = – 20°
Thus, Monday, a = t₁ = – 20°
Using formula t_{n + 1} = t_n + d
We can find the value of the other terms
Tuesday, t₂ = t₁ + d = – 20 – 5 = – 25°
Wednesday, t₃ = t₂ + d = – 25 – 5 = – 30°
Thursday, t₄ = t₃ + d = – 30 – 5 = – 35°
Friday, t₅ = t₄ + d = – 35 – 5 = 40°
Saturday, t₆ = t₅ + d = – 40 – 5 = – 45°
Thus, we obtain an A.P.
– 20°, – 25°, – 30°, – 35°, – 40°, – 45°

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Question 414 Marks

There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the 15th row and also find how many total seats are there in the auditorium?

Answer

Given: first term a = 20
Second term t₁ = 22
Third term t₂ = 24
Common difference d = t₂ – t₁ = 24 – 22 = 2
We need to find t₁₅ thus n = 15
Now, By using n^th term of an A.P. formula
t_n = a + (n – 1)d
where n = no. of terms
a = first term
d = common difference
t_n = n^th terms
On substituting all value in n^th term of an A.P.
⇒ t₁₅ = 20 + (15 – 1) × 2
⇒ t₁₅ = 20 + 14 × 2
⇒ t₁₅ = 20 + 28 = 48
We have been given that, there are 27 rows in an auditorium
Thus, we need to find total seats in auditorium i.e. S₂₇Now, By using sum of n^th term of an A.P. we will find it’s sum
$S_n=\frac{n}{2}[2 a+(n-1) d]$
Where, $n=n o$. of terms
$\begin{array}{l}
a=\text { first term } \\
d=\text { common difference } \\
S_n=\text { sum of } n \text { terms }
\end{array}$
Thus, on substituting the given value in formula we get,
$\begin{array}{l}
\Rightarrow S_{27}=\frac{27}{2}[2 \times 20+(27-1) \times 2] \\
\Rightarrow S_{27}=\frac{27}{2} \times 2 \times[20+26] \\
\Rightarrow S_{27}=27 \times 46 \\
\Rightarrow S_{27}=1242
\end{array}$

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Question 424 Marks

Sachin invested ina national saving certificate scheme. In the first year he invested ₹ 5000, in the second year ₹ 7000, in the third year ₹ 9000 and so on. Find the total amount that he invested in 12 years.

Answer

By given information we can form an A.P.
5000, 7000,9000 , $\qquad$
Hence, the first term $a =5000$
Second term $t_1=7000$
Third term $t_2=9000$
Thus, common difference $d=t_2-t_1=9000-7000=2000$
Here, number of terms $n =12$
We need to find $S_{12}$
Now, By using sum of $n^{\text {th }}$ term of an A.P. we will find it's sum
$S_n=\frac{n}{2}[2 a+(n-1) d]$
Where, $n=$ no. of terms
$a=$ first term
$d=$ common difference
$S_n=$ sum of $n$ terms
Thus, on substituting the given value in formula we get,
$\begin{array}{l}\Rightarrow S_{12}=\frac{12}{2}[2 \times 5000+(12-1) \times 2000] \\ \Rightarrow S_{12}=6 \times[10,000+11 \times 2000] \\ \Rightarrow S_{12}=6 \times[10,000+22,000] \\ \Rightarrow S_{12}=6 \times 32,000 \\ \Rightarrow S_{12}=\text { Rs. } 192000\end{array}$

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Question 434 Marks

A man borrows ₹ 8000 and agrees to repay with a total interest of ₹ 1360 in 12 monthly instalments. Each instalment being less than the preceding one by ₹ 40. Find the amount of the first and last instalment.

Answer

Given: A man borrows $=$ Rs. 8000
Repay with total interest $=$ Rs 1360
In 12 months, thus $n =12$
Thus, $S_{12}=8000+1360=9360$
Each installment being less than preceding one
Thus, $d=-40$
We need to find "a"
Now, By using sum of $n^{\text {th }}$ term of an A.P. we will find it's sum
$S_n=\frac{n}{2}[2 a+(n-1) d]$
Where, $n=$ no. of terms
$a=\text { first term }$
$d=$ common difference
$S _n=$ sum of $n$ terms
Thus, on substituting the given value in formula we get,
$\Rightarrow S _{12}=\frac{12}{2}[2 a +(12-1) \times(-40)]$
$\begin{array}{l}
\Rightarrow 9360=6[2 a-11 \times 40] \\
\Rightarrow \frac{9360}{6}=2 a-440 \\
\Rightarrow 1560=2 a-440 \\
\Rightarrow 1560+440=2 a \\
\Rightarrow 2 a=2000 \\
\Rightarrow a=\frac{2000}{2}=1000
\end{array}$
Thus, first installment $a =$ Rs. 1000
Now, By using sum of $n^{\text {th }}$ term of an A.P. we will find it's sum
$S _{ n }=\frac{ n }{2}[\text { first term }+ \text { last term }]$
Where, $n=$ no. of terms
$S_n=\text { sum of } n \text { terms }$
Thus, on substituting the given value in formula we get,
Let $a=$ first term, $t_n=$ last term
$\Rightarrow S _{12}=\frac{12}{2}\left[ a + t _{ n }\right]$
$\begin{array}{l}
\Rightarrow 9360=6\left[1000+t_n\right] \\
\Rightarrow 1000+t_n=\frac{9360}{6}=1560 \\
\Rightarrow t_n=1560-1000=560
\end{array}$
Thus, last installment $t_n=560$

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Question 444 Marks

On 1^st Jan 2016, Sanika decides to save ₹ 10, ₹ 11 on second day, ₹ 12 on third day. If she decides to save like this, then on 31st Dec 2016 what would be her total saving?

Answer

By given information we can form an A.P.
$10,11,12,13, \ldots \ldots$
Hence, the first term $a=10$
Second term $t _1=11$
Third term $t _2=12$
Thus, common difference $d=t_2-t_1=12-11=1$
Here, number of terms from $1^{\text {st }}$ Jan 2016 to $31^{\text {st }}$ Dec 2016 is, $n =366$
We need to find $S _{366}$
Now, By using sum of $n ^{\text {th }}$ term of an A.P. we will find it's sum
$S _{ n }=\frac{ n }{2}[2 a +( n -1) d ]$
Where, $n=$ no. of terms
$\begin{array}{l}
a=\text { first term } \\
d=\text { common difference } \\
S_n=\text { sum of } n \text { terms }
\end{array}$
Thus, on substituting the given value in formula we get,
$\begin{array}{l}
\Rightarrow S_{366}=\frac{366}{2}[2 \times 10+(366-1) \times 1] \\
\Rightarrow S_{366}=183[20+365] \\
\Rightarrow S_{366}=183 \times 385 \\
\Rightarrow S_{366}=\operatorname{Rs} 70,455
\end{array}$

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Question 454 Marks

On the world environment day tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.

Answer

First term a = 1
Second term t₁ = 2
Third term t₃ = 3
Common difference d = t₃ – t₂ = 3 – 2 = 1
We need to find total number of trees when n = 25
Thus, By using sum of n^th term of an A.P. we will find it’s sum
$S_n=\frac{n}{2}[2 a+(n-1) d]$
Where, $n=$ no. of terms
$\begin{array}{l}
a=\text { first term } \\
d=\text { common difference } \\
S_n=\text { sum of } n \text { terms }
\end{array}$
We need to find $S_{25}$
Thus, on substituting the given value in formula we get,
$\begin{array}{l}
\Rightarrow S_{25}=\frac{25}{2}[2 \times 1+(25-1) \times 1] \\
\Rightarrow S_{25}=\frac{25}{2}[2+24] \\
\Rightarrow S_{25}=\frac{25}{2} \times 2 \times[1+12] \\
\Rightarrow S_{25}=25 \times 13=325
\end{array}$

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Question 464 Marks

If the 9^th term of an A.P. is zero then show that the 29^th term is twice the 19^th term.

Answer

Now, By using n^th term of an A.P. formulat_n = a + (n – 1)d
where n = no. of terms
a = first term
d = common difference
t_n = n^th terms
Given: t₉ = 0
⇒ t₉ = a + (9 – 1)d
⇒ 0 = a + 8d
⇒ a = – 8d
To Show: t₂₉ = 2× t₁₉
Now,
⇒ t₂₉ = a + (29 – 1)d
⇒ t₂₉ = a + 28d
⇒ t₂₉ = – 8d + 28d = 20 d (since, a = – 8d )
⇒ t₂₉ = 20 d
⇒ t₂₉ = 2 × 10 d ….(1)
Also,
⇒ t₁₉ = a + (19 – 1)d
⇒ t₁₉ = a + 18d
⇒ t₁₉ = – 8d + 18d = 10 d (since, a = – 8d )
⇒ t₁₉ = 10 d …..(2)
From eq. (1) and eq. (2) we get,
t₂₉ = 2× t₁₉

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Question 474 Marks

Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14.
(Assume the four consecutive terms in A.P. are a – d, a, a + d, a + 2d.)

Answer

$\begin{array}{l}
\text { Let the first term be } a-d \\
\text { the second term be } a \\
\text { the third term be } a+d \\
\text { the fourth term be } a+2 d \\
\text { Given: sum of consecutive four term is } 12 \\
\Rightarrow(a-d)+a+(a+d)+(a+2 d)=12 \\
\Rightarrow 4 a+2 d=12 \\
\Rightarrow 2(2 a+d)=12 \\
\Rightarrow 2 a+d=\frac{12}{2}=6 \\
\Rightarrow 2 a+d=6 \ldots . .(1)
\end{array}$
Also, sum of third and fourth term is 14
$\begin{array}{l}
\Rightarrow(a+d)+(a+2 d)=14 \\
\Rightarrow 2 a+3 d=14 \ldots . .(2)
\end{array}$
Subtracting eq. (1) from eq. (2) we get,
$\begin{array}{l}
\Rightarrow(2 a+3 d)-(2 a+d)=14-6 \\
\Rightarrow 2 a+3 d-2 a-d=8
\end{array}$
$\begin{array}{l}
\Rightarrow 2 d=8 \\
\Rightarrow d=\frac{8}{2}=4 \\
\Rightarrow d=4
\end{array}$
Substituting value of "d" in eq. (1) we get,
$\begin{array}{l}
\Rightarrow 2 a+4=6 \\
\Rightarrow 2 a=6-4=2 \\
\Rightarrow a=\frac{2}{2}=1 \\
\Rightarrow a=1
\end{array}$
Thus, $a=1$ and $d=4$
Hence, first term $a-d=1-4=-3$
the second term $a=1$
the third term $a+d=1+4=5$
the fourth term $a +2 d =1+2 \times 4=1+8=9$
Thus, the A.P. is $-3,1,5,9$

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Question 484 Marks

In an A.P. sum of three consecutive terms is 27 and their product is 504 find the terms? (Assume that three consecutive terms in A.P. are a – d, a, a + d.)

Answer

Let the first term be a – d
the second term be a
the third term be a + d
Given: sum of consecutive three term is 27
⇒ (a – d) + a + (a + d) = 27
⇒ 3 a = 27
$\Rightarrow a =\frac{27}{3}=9$
Also, Given product of three consecutive term is 504
$\begin{array}{l}
\Rightarrow(a-d) \times a \times(a+d)=504 \\
\Rightarrow(9-d) \times 9 \times(9+d)=504(\text { since, } a=9) \\
\Rightarrow(9-d) \times(9+d)=\frac{504}{9}=56
\end{array}$
⇒ 9² – d² = 56 (since, (a – b)(a + b) = a² – b²)
⇒ 81 – d² = 56
⇒ d² = 81 – 56 = 25
⇒ d = √25 = ± 5
Case 1:
Thus, if a = 9 and d = 5
Then the three terms are,
First term a – d = 9 – 5 = 4
Second term a = 9
Third term a + d = 9 + 5 = 14
Thus, the A.P. is 4, 9, 14
Case 2:
Thus, if a = 9 and d = – 5
Then the three terms are,
First term a – d = 9 – ( – 5) = 9 + 5 = 14
Second term a = 9
Third term a + d = 9 + ( – 5) = 9 – 5 = 4
Thus, the A.P. is 14, 9, 4

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Question 494 Marks

Sum of first 55 terms in an A.P. is 3300, find its 28th term.

Answer

Given: $S_{55}=3300$ where $n=55$
Now, By using sum of $n^{\text {th }}$ term of an A.P. we will find it's sum
$S_n=\frac{n}{2}[2 a+(n-1) d]$
Where, $n=$ no. of terms
$a =$ first term
$d=$ common difference
$S _{ n }=$ sum of $n$ terms
Thus, on substituting the given value in formula we get,
$\begin{array}{l}
\Rightarrow S_{55}=\frac{55}{2}[2 a+(55-1) d] \\
\Rightarrow 3300=\frac{55}{2}[2 a+54 d] \\
\Rightarrow 3300=\frac{55}{2} \times 2 \times[a+27 d]
\end{array}$
$\begin{array}{l}
\Rightarrow 3300=55 \times[a+27 d] \\
\Rightarrow \frac{3300}{55}=a+27 d \\
\Rightarrow a+27 d=60 \ldots \ldots .(1)
\end{array}$
We need to find value of $28^{\text {th }}$ term i.e $t_{28}$
Now, By using $n^{\text {th }}$ term of an A.P. formula
$t_n=a+(n-1) d$
where $n=n o$. of terms
$\begin{array}{l}
a=\text { first term } \\
d=\text { common difference } \\
t_n=n^{\text {th }} \text { terms }
\end{array}$
we can find value of $t_{28}$ by substituting all the value in formula we get,
⇒ t₂₈ = a + (28 – 1) d
⇒ t₂₈ = a + 27 d
From eq. (1) we get,
⇒ t₂₈ = a + 27 d = 60
⇒ t₂₈ = 60

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Question 504 Marks

In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.

Answer

Given: t₁₉ = 52 and t₃₈ = 128
To find: value of “a” and “d”
Using n^th term of an A.P. formula
t_n = a + (n – 1)d
where n = no. of terms
a = first term
d = common difference
t_n = n^th terms
we will find value of “a” and “d”
Let, t₁₉ = a + (19 – 1) d
⇒ 52 = a + 18 d …..(1)
t₃₈ = a + (38 – 1) d
⇒ 128 = a + 37 d …..(2)
Subtracting eq. (1) from eq. (2), we get,
⇒ 128 – 52 = (a – a) + (37 d – 18 d)
⇒ 76 = 19 d
$\Rightarrow d=\frac{76}{19}=4$
Substitute value of “d” in eq. (1) to get value of “a”
⇒ 52 = a + 18 ×4
⇒ 52 = a + 72
⇒ a = 52 – 72 = – 20
Now, to find value of S₅₆ we will using formula of sum of n terms
$S_n=\frac{n}{2}[2 a+(n-1) d]$
Where, n = no. of terms
a = first term
d = common difference
S_n = sum of n terms
Thus, Substituting given value in formula we can find the value of S_n$\Rightarrow S_{56}=\frac{56}{2}[2 \times(-20)+(56-1) \times 4]$
⇒S₅₆ = 28 × [ – 40 + 55×4]
⇒S₅₆ = 28 × [ – 40 + 220]
⇒S₅₆ = 28 × 180 = 5040
Thus, S₅₆ = 5040

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Question 514 Marks

Find the sum of all even numbers from 1 to 350.

Answer

List of even natural number between 1 to 350 is
2,4,6,…….348
Where first term a = 2
Second term t₁ = 4
Third term t₂ = 6
Thus, common difference d = t₂ – t₁ = 6 – 4 = 2
t_n = 348 (As we have to find the sum of even numbers between 1 and 350 therefore excluding 350 )
Now, By using n^th term of an A.P. formula
t_n = a + (n – 1)d
where n = no. of terms
a = first term
d = common difference
t_n = n^th terms
we can find value of “n” by substituting all the value in formula we get,
⇒ 348 = 2 + (n – 1) × 2
⇒ 348 – 2 = 2(n – 1)
⇒ 346 = 2(n – 1)
$\Rightarrow n -1=\frac{346}{2}=173$
⇒ n = 173 + 1 = 174
Now, By using sum of n^th term of an A.P. we will find it’s sum
$S_n=\frac{n}{2}[2 a+(n-1) d]$
Where, n = no. of terms
a = first term
d = common difference
S_n = sum of n terms
Thus, Substituting given value in formula we can find the value of S_n$\begin{array}{l}\Rightarrow S_{174}=\frac{174}{2}[2 \times 2+(174-1) \times 2] \\ \Rightarrow S_{174}=\frac{174}{2}[4+173 \times 2] \\ \Rightarrow S_{174}=\frac{174}{2}[4+346] \\ \Rightarrow S_{174}=\frac{174}{2} \times 350=174 \times 175=30,450\end{array}$
Thus, S₁₇₄ = 30,450

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Question 524 Marks

Find the sum of first 123 even natural numbers.

Answer

List of first 123 even natural number is
$2,4,6, \ldots \ldots$
Where first term $a =2$
Second term $t _1=4$
Third term $t _2=6$
Thus, common difference $d=t_2-t_1=6-4=2$
$n=123$
By using sum of $n^{\text {th }}$ term of an A.P. is
$S_n=\frac{n}{2}[2 a+(n-1) d]$
Where, $n=$ no. of terms
$a =$ first term
$d =$ common difference
$S_n=$ sum of $n$ terms
Thus, Substituting given value in formula we can find the value of $S_n$
$\begin{array}{l}
\Rightarrow S_n=\frac{123}{2}[2 \times 2+(123-1) \times 2] \\
\Rightarrow S_n=\frac{123}{2}[4+122 \times 2] \\
\Rightarrow S_n=\frac{123}{2}[4+244] \\
\Rightarrow S_n=\frac{123}{2} \times 248=123 \times 122=15252
\end{array}$
Thus, $S_n=15252$

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Question 534 Marks

Answer

Semi circumferences $P _1, P _2, P _3, \ldots$ are drawn by taking centres $A , B , A$, $B , \ldots$ It is given that radius of the first circle is $0.5 cm$. The radius of the second circle is $1.0 cm , \ldots$ From this information we will find $P _1, P _2, P _3, \ldots P _3$.
Length of the first semi circumference $=P_1=\pi r_1=\pi \times \frac{1}{2}=\frac{\pi}{2}$
$\begin{array}{l}
P_2=\pi r_2=\pi \times 1=\pi \\
P_3=\pi r_3=\pi \times 1.5=\frac{3}{2} \pi
\end{array}$
The lengths are $P_1, P_2, P_3, \ldots$, and the numbers $\frac{1}{2} \pi, 1 \pi, \frac{3}{2} \pi, \ldots$ are in A.P. Here $a=\frac{1}{2} \pi, d=\frac{1}{2} \pi$, From this let's find $S _{13}$.
$\begin{aligned}
S _{ n } & =\frac{n}{2}[2 a+(n-1) d] \\
S _{13} & =\frac{13}{2}\left[2 \times \frac{\pi}{2}+(13-1) \times \frac{1}{2} \pi\right] \\
& =\frac{13}{2}[\pi+6 \pi] \\
& =\frac{13}{2} \times 7 \pi= \\
& =\frac{13}{2} \times 7 \times \frac{22}{7} \\
& =143 cm .
\end{aligned}$
$\therefore$ The total length of spiral shape formed by 13 semicircles is $143 cm$.

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Question 544 Marks

Anvar saves some amount every month.In first three months he saves ₹ 200, ₹ 250 and ₹ 300 respectively. In which month will he save ₹ 1000?

Answer

Saving in first month ₹ 200 ; Saving in second month $₹ 250 ; \ldots$. $200,250,300, \ldots$ this is an A.P.
Here $a=200, d=50$, Let's find $n$ using $t _{ n }$ formula and then find $S _{ n }$.
$\begin{aligned}
t _{ n } & =a+(n-1) d \\
& =200+(n-1) 50 \\
& =200+50 n-50 \\
1000 & =150+50 n \\
150+50 n & =1000 \\
50 n & =1000-150 \\
50 n & =850 \\
\therefore n & =17
\end{aligned}$
In the $17^{\text {th }}$ month he will save $₹ 1000$.
Let's find that in 17 months how much total amount is saved.
$\begin{aligned}
S _{ n } & =\frac{n}{2}[2 a+( n -1) d] \\
& =\frac{17}{2}[2 \times 200+(17-1) \times 50] \\
& =\frac{17}{2}[400+800] \\
& =\frac{17}{2}[1200] \\
& =17 \times 600 \\
& =10200
\end{aligned}$
In 17 months total saving is ₹ 10200 .

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Question 554 Marks

Ajay sharma repays the borrowed amount of ₹ 3,25,000 by paying ₹ 30500 in the first month and then decreases the payment by ₹ 1500 every month. How long will it take to clear his amount?

Answer

Let the time required to clear the amount be $n$ months. The monthly payment decreases by ₹ 1500 . Therefore the payments are in A.P.
$\begin{array}{l}
\text { First term }=a=30500, d=-1500 \\
\text { Amount }= S _{ n }=3,25,000 \\
S _{ n }=\frac{n}{2}[2 a+(n-1) d] \\
3,25,000=\frac{n}{2}[2 \times 30500+(n-1) d] \\
\quad=\frac{n}{2}[2 \times 30500-1500 n+1500] \\
3,25,000=30500 n-750 n^2+750 n \\
\therefore 750 n^2-31250 n+325000=0
\end{array}$
divide both sides by 250 .
$\begin{array}{l}
\therefore 3 n^2-125 n+1300=0 \\
\therefore 3 n^2-60 n-65 n+1300=0 \\
\therefore 3 n(n-20)-65(n-20)=0 \\
\therefore(n-20)(3 n-65)=0 \\
\therefore n-20=0,3 n-65=0 \\
\therefore n=20 \text { or } n=\frac{65}{3}=21 \frac{2}{3}
\end{array}$
In an A.P. $n$ is a natural number.
$\therefore n \neq \frac{65}{3} \quad \therefore \quad n=20$
(Or, after 20 months, $S _{20}=3,25,000$ then the total amount will be repaid. It is not required to think about further period of time.)
$\therefore$ To clear the amount 20 months are needed.

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Question 564 Marks

A mixer manufacturing company manufactured 600 mixers in 3^rd year and in 7^th year they manufactured 700 mixers. If every year there is same growth in the production of mixers then find (i) Production in the first year (ii) Production in 10^th year (iii) Total production in first seven years.

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Question 574 Marks

The 10^th term and the 18^th term of an A.P. are 25 and 41 respectively then find 38^th term of that A.P., similarly if n^th term is 99. Find the value of n.

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Question 584 Marks

Neeta saves in a 'Mahila Bachat Gat' ₹ 2 on the first day, ₹ 4 on the second day, ₹ 6 on the third day and so on. What will be her saving in the month of February 2010 ?

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Question 594 Marks

Mr. Shah borrows ₹ 4000 and agrees to repay with a total interest of ₹ 500 in 10 installments, each installment being less than the preceding installment by ₹ 10. What should be the first and the last installment?

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Question 604 Marks

Jinal saves ₹ 1600 during the first year, ₹ 2100 in the second year, ₹ 2600 in the third year, If she continues her saving in this pattern, in how many years will she save ₹ 38,500 ?

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Question 614 Marks

In winter, the temperature at a hill station from Monday to Friday is in A.P., The sum of the temperatures of Monday, Tuesday and Wednesday is zero and the sum of the temperature of Thursday and Friday is 15. Find the temperature of each of the five days.

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Question 624 Marks

If sum of $m$ terms is $n$ and sum of $n$ terms is $m$, then show that the sum of $(m+n)$ terms is $-(m+n)$.

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Question 634 Marks

If $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of an A. P. are $l, m, n$ respectively, show that
$(q-r) l+(r-p) m+(p-q) n=0$

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Question 644 Marks

How many terms of the A.P. 16, 14, 12, ..... Are needed to given the sum 60? Explain why we get two answers.

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Question 654 Marks

Find the middle term of sequence formed by all three digit numbers which leave a remainder 3 when divided by 4. Also find sum of all numbers on both sides of the middle term.

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Question 664 Marks

A meeting hall has 20 seats in the first row, 20 seats in the second, 28 seats in the third row and so on and has in all 30 rows. How many seats are there in the meeting hall?

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Question 674 Marks

A farmer borrows ₹ 1000 and agrees to repay with a total interest of ₹ 140 in 12 installments, each installment being less than the preceding installment by ₹ 10 . What should be his first installment?

Answer

₹ 150

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Question 684 Marks

200 logs of wood are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows 200 logs are placed and how many logs are there in the top row?

Answer

The arrangement of logs $20,19,18, \ldots$. forms an A.P. with $a=20, d=-1$
Let 200 logs be arranged in n rows.
$S_n=200$
$S_n=\frac{n}{2}[2 a+(n-1) d]$
$\therefore \quad 200=\frac{n}{2}[2 \times 20+(n-1)(-1)]$
$\therefore \quad 400=n[40-n+1]$
$\therefore \quad 400=n[41-n]$
$\therefore \quad 400=41 n-n^2$
$\therefore \quad n^2-41 n+400=0$
$\therefore \quad n^2-25 n-16 n+400=0$
$\therefore n(n-25)-16(n-25)=0$
$\therefore \quad(n-25)(n-16)=0$
$\therefore n-25=0$ or $n-16=0$
$\therefore n=25$ or $n=16$
If $n=25$
$t_n=a+(n-1) d$
$\therefore \quad t_{25}=20+(25-1)(-1)$
$\begin{array}{l}=20+24(-1) \\ =20-24\end{array}$
$t_{25}=-4$
No. of logs in the $25^{ th }$ row cannot be negative
$\therefore n \neq 25 \quad \therefore n=16$
$t_n=a+(n-1) d$
$\begin{aligned} \therefore \quad t_{16} =20+(16-1)(-1) \\ =20+15(-1) \\ =20-15 \\ t_{16} =5\end{aligned}$
$\therefore$ 200 logs are placed in 16 rows and there are 5 logs in the top row.

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