Answer the above questions
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Answers
Step-by-step explanation:
In Ap term - it's previous term is constant, so 3k-(k+1) =(4k+2) = slove the equation ,3k-k-1=4k+2 =k= -3/2
Answer:
1. d
2.d
3.c
4.a
5.c
6.b
7.b
8.b
9.d
10.d
Step-by-step explanation:
1)The three consecutive terms in an AP are (k+1) ,3k and (4k+2)
We know that
If a,b,c are the three consecutive terms in an AP then b = (a+c)/2
now ,
Let a = k+1 , b = 3k and c = 4k+2
=> 3k = (k+1+4k+2)/2
=> 3k = (5k+3)/2
=> 2×3k = 5k+3
=>6k = 5k+3
=> 6k-5k = 3
=> k = 3
The value of k = 3
2)Given terms in an AP are √2,√8,√18
√8 = √(2×2×2) = 2√2
√18 = √(2×3×3) = 3√2
The AP : √2,2√2,3√2
next term = 4√2> √(4×4×2)
=> √32
The next term of the AP = √32
3)The first term of an AP = p
The common difference = q
We know that
nth term of an AP = a+(n-1)d
10th term = a+(10-1)d = a+9d
t10 = p+9q
Tenth term of the AP is p+9q
4)Given AP is -3 ,-1/2,2,..
First term = -3
Common difference = (-1/2)-(-3)
=> d = (-1/2)+3
=>d = (-1+6)/2
=> d = 5/2
We know that nth term = an = a+(n-1)d
11th term = (-3)+(11-1)(5/2)
=> a 11 = -3+(10)(5/2)
=> a 11 = -3+(5×5)
=> a 11 = -3+25
=> a 11 = 22
11th term of the AP = 22
5)Given AP is 21 ,18,15,...
First term = 21
Common difference = 18-21 = -3
Let an = -81
We know that
an = a+(n-1)d
=> 21+(n-1)(-3) = -81
=> 21-3n+3 = -81
=> 24-3n = -81
=> 24+81 = 3n
=> 105= 3n
=> n = 105/3
=> n = 35
35th term of the AP is -81
6) Given nth term of an AP = 7-4n
Put n = 1 then t1 = 7-4(1) = 7-4 = 3
Put n = 2 then t2 =7-4(2) = 7-8 = -1
Common difference = t2-t1 = -1-3 = -4
7)Given AP is 21 ,42,63,84,...
First term = 21
Common difference = 42-21 = 21
Let an = 210
We know that
an = a+(n-1)d
=> 21+(n-1)(21) = 210
=> 21+21n-21= 210
=> 21n = 210
=> n = 210/21
=> n =10
10th term of the AP is 210
8) Given A.P = 3,8,13,...,253
Common difference = 8-3 = 5
The AP from the end = 253, 248,245,...,13,8,3
First term from the end = 253
Common difference = 248-253 = -5
We know that
an = a+(n-1)d
t20 = 253+(20-1)(-5)
=> t 20 = 253 +(19)(-5)
=> t 20 = 253-95
=> t 20 = 158
20th term from the end of the AP = 158
9)Given that
5+13+21+...+181
First term = 5
Common difference = 13-5 = 8
Last term = an = 181
=> a+(n-1)d = 181
=> 5+(n-1)(8) = 181
=> 5+8n-8 = 181
=> 8n-3 = 181
=> 8n = 181+3
=> 8n = 184
=> n = 184/8
=> n = 23
Sum of the first n terms = Sn = (n/2)(a+an)
Sum of 23 terms = S 23
=> (23/2)(5+181)
=> (23/2)(186)
=> 23×93
=>2139
10) Given AP :√2,√8,√18,√32,...
First term = √2
Common difference = √8-√2
=> √(2×2×2)-√2
=> 2√2-√2
=> √2
We know that
Sum of first n terms of an AP
=> Sn = (n/2)[2a+(n-1)d]
=> Sn = (n/2)[2√2+(n-1)(√2)]
=> Sn = (n/2)[2√2+√2 n -√2]
=> Sn = (n/2)[√2 n +√2)
=> Sn = (n/2)×(√2)(n+1)
=> Sn = (n/√2)(n+1)
=> Sn = n(n+1)/√2
Used formulae:-
- nth term of an AP = a+(n-1)d
- Sum of first n terms of an AP
- Sn = (n/2)[2a+(n-1)d]
- Sum of the first n terms = Sn =
(n/2)(a+an)
- If a,b,c are the three consecutive terms in an AP then b = (a+c)/2