a. True ,d=5
b. True ,d=0
c. False
d. True ,d=11
f. True ,d=-3
Next two terms are -11,-13
T4=a+3d
T6=a+5d
T8=a+7d
T10=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
T1=a=−13
T2=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
Sn−1=3(n−1)2−4(n−1)
=3n2+3−6n−4n+4=3n2−10n+7
Nth will be
Tn=Sn−Sn−1
=3n2−4n−3n2+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
Sn=287
n2[2a+(n−1)d]=287
n2[2+3(n−1)]=287
3n2−n−574=0
Solving this Quadratic equation
using quadratic formula
n=−b±√b2−4ac2a
n= 14 or -41/3
So n=14 as it cannot be negative
Now
x=a+(n−1)d
x=1+13×3=40
a18−a13=a+17d−(a−12d)=5d=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=2x−yx+y
S11=112[2x−yx+y+(11−1)2x−yx+y
=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
T8=−11+(8−1)4=17
T9=−11+(9−1)4=21
Let 1st term be a and common difference =d
According to the question,
T4=3T1
a+3d=3a
3d=2a -(1)
Also,
T7=2T3+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+ 5nS_1 = 3(1)^2 + 5(1)�
= 3 + 5=8Thereforea_1 = 8S_2 = 3(2)^2 + 5(2)
= 12 + 10
= 22
Thereforea_1+ a_2 =22
8 + a_2 = 22a_2 = 14So,theAPis8,14......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
an=a1+(n−1)×d
490=10+(n�1)×10
n=49
So, sum of all three digit numbers which are divisible by 7
S49=n2[a1+an]=492[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=S2+S5−S10
Now Sum of number multiples of 2
S2=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
an=a1+(n−1)×d
500=2+(n−1)×2
n=250
Then,
S2=2+4+6....+500=n2[a1+an]=2502[2+500]=62750
Now Sum of number multiples of 5
S5=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
an=a1+(n−1)×d
500=5+(n−1)×5
n=100
Then,
S5=5+10+15....+500=n2[a1+an]=1002[5+500]=25250
Now Sum of number multiples of 10
S10=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
an=a1+(n−1)×d
500=10+(n−1)×10
n=50
Then,
S5=10+20+30....+500=n2[a1+an]=502[10+500]=12750
Therefore,
S=S2+S5−S10
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
an=a1+(n−1)×d
998=103+(n−1)×5
n=180
Therefore
S=103+108+111...+998=n2[a1+an]=1802[103+998]=99090
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