a. True ,d=5
b. True ,d=0
c. False
d. True ,d=11
f. True ,d=-3
Next two terms are -11,-13
$T_4=a+3d$
$T_6=a + 5d$
$T_
8=a+7d$
$T_{10}=a+9d$
As per the question
$a+3d+a+7d=24$
or
$2a+10d=24$ -(1)
$a+5d+a+9d=44$
$2a+14d=44$ -(2)
Solving 1 and 2
a=-13, d=5
Now
$T_1=a=-13$
$T_2=a+d=-13+(5)=-8$
$T=a+2d=-13+2(5)=-13+10=-3$
a. True
b. True
c. False
d. True
e. True
Option (b) is correct
$7 - (2a-1) = 3a -7$
$15=5a$
$a=3$
Option (c) is correct
The AP will be 12, 16 ... 248
Now
$248 = 12 + (n-1) 4$
$n=60$
Option (d) is correct
$(2p+5) - (3p+7) = (2p+7) - (2p+5)$
$-p -2=2$
$p=-4$
Option (c) is correct
$S_{n-1} = 3(n-1)^2 - 4(n-1)$
$=3n^2 +3 -6n -4n +4=3n^2 -10n + 7$
Nth will be
$T_n =S_n - S_{n-1}$
$= 3n^2 - 4n -3n^2 +10n - 7=6n -7$
Option (a) is correct
(10000)
Option (a) is correct
Let there be n term in the series in the sum
Now a=1 ,d=3
$1 + 4 + 7 + 10 +...+ x =287$
$S_n=287$
$\frac {n}{2} [2a +(n-1)d]=287$
$\frac {n}{2} [2 + 3(n-1)]=287$
$3n^2 -n - 574 = 0$
Solving this Quadratic equation
using quadratic formula
$n= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
n= 14 or -41/3
So n=14 as it cannot be negative
Now
$x= a + (n-1) d$
$x= 1 + 13 \times 3 = 40$
$a_{18} -a_{13}= a+ 17d - (a - 12 d) = 5 d= 25$
As n � 2, 4n � 1, 5n + 2 are in AP,
so (4n � 1) � (n � 2) = (5n + 2) � (4n � 1)
i.e, 3n + 1 = n + 3
i.e, n = 1
$d=\frac {2x -y}{x+ y}$
$S_{11} = \frac {11}{2} [ \frac 2{x- y}{x+y} + (11 -1) \frac {2x -y}{x+ y}$
$=\frac {11(11x- 6y)}{x+y}$
Here a=-11, d=4
Now
$49=-11 + (n-1)4$
n=16
So middle most terms will 8 and 9
$T_8= -11 + (8-1)4=17$
$T_9=-11 + (9-1)4=21$
Let 1st term be a and common difference =d
According to the question,
$T_4 =3T_1$
$a + 3d = 3a$
$3d = 2a$ -(1)
Also,
$T_7 =2T_3 + 1$
$a + 6d = 2[a + (2d] + 1$
$2d = a+1$ -(2)
Solving equation (1) and (2)
a = 3,d=2
So, the AP formed is 3, 5, 7, 9, .........
Given: S_n = 3n^2+ 5n$
$S_1 = 3(1)^2 + 5(1)�
= 3 + 5=8$
Therefore
$a_1 = 8$
$S_2 = 3(2)^2 + 5(2)
= 12 + 10
= 22
$
Therefore
$a_1+ a_2 =22$
$
8 + a_2 = 22$
$a_2 = 14$
So, the AP is
8 ,14......
$a_16= a_1 +(n-1)d = 8 + 15 \times 6 = 98$
Let the angles be a-d,a,a+d
According to the question
$a+d = 2(a-d)$
or
$a=3d$ -(1)
Since sum of the angles are 180 degrees
$a-d+a+a+d = 180$
$3a = 180$
a = 60
From Ist equation
$3d = 60$
$
d = 20$
so the angles are 40°, 60° and 80
°
i. LCM of 2 and 5 =10.
Now all those integers which are multiples of 10 are also the multiples of 2 and 5.
So number between 1 and 500 will be given as
10, 20, 30, ...., 490�
Sum = 10 + 20 + 30..... + 490
Now this series is a AP with first term as 10, common difference=10, last term =490, n=?
Now from nth term formula
$a_n= a_1 + (n-1) \times d$
$490 = 10 + (n � 1) \times 10$
$n = 49$
So, sum of all three digit numbers which are divisible by 7
$S_{49} = \frac {n}{2} [a_1 + a_n] = \frac {49}{2} [10 + 490]= 12250$
ii.
This question can be divided as
S= Sum of number multiples of 2 + Sum of number multiples of 5 - Sum of number multiples of 2 and 5 both
S=Sum of number multiples of 2 + Sum of number multiples of 5 - Sum of number multiples of 10
$S= S_2 + S_5 - S_{10}$
Now Sum of number multiples of 2
$S_2 =2 + 4 + 6....+ 500$
This is a AP with first term =2 and common difference =2,last term=500
Number of term can found from nth term formula
$a_n= a_1 + (n-1) \times d$
$500 = 2 + (n-1) \times 2$
n=250
Then,
$S_2 =2 + 4 + 6....+ 500 = \frac {n}{2} [a_1 + a_n] = \frac {250}{2} [2 + 500]=62750$
Now Sum of number multiples of 5
$S_5 =5 + 10 + 15....+ 500$
This is a AP with first term =5 and common difference =5,last term=500
Number of term can found from nth term formula
$a_n= a_1 + (n-1) \times d$
$500 = 5 + (n-1) \times 5$
n=100
Then,
$S_5 =5 + 10 + 15....+ 500= \frac {n}{2} [a_1 + a_n] = \frac {100}{2} [5 + 500]=25250$
Now Sum of number multiples of 10
$S_{10} =10 + 20 + 35....+ 500$
This is a AP with first term =10 and common difference =10,last term=500
Number of term can found from nth term formula
$a_n= a_1 + (n-1) \times d$
$500 = 10 + (n-1) \times 10$
n=50
Then,
$S_5 =10 + 20 + 30....+ 500= \frac {n}{2} [a_1 + a_n] = \frac {50}{2} [10 + 500]=12750$
Therefore,
$S= S_2 + S_5 - S_{10}$
$S= 62750+25250-12750$
$S= 75750$
The smallest 3 digit no. = 100 and greatest 3 digit number is 999
Since 100 is divisible by 5,adding 3 on 100 will provide the number which leave the remainder 3 when divided by 5
So the smallest three digit number which is divisible by 5 and gives reminder 3= 103
The largest 3 digit no ,
999/5 gives reminder 4 ,So subtracting 1 will the number which leave the remainder 3 when divided by 5
So the Largest three digit number which is divisible by 5 and gives reminder 3= 998
Similarly we can find other numbers, the number will be given as
103,108,111,...998
Now $S= 103+ 108 + 111... + 998$
This is a AP with first term =103 and common difference =5,last term=998
Number of term can found from nth term formula
$a_n= a_1 + (n-1) \times d$
$998 = 103 + (n-1) \times 5$
n=180
Therefore
$S= 103+ 108 + 111... + 998 = \frac {n}{2} [a_1 + a_n] = \frac {180}{2} [103 + 998]=99090$
This Class 10 Maths Worksheet for Arithmetic Progression with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.You can download in PDF form also using the below links