Question

1. The nth term of an A.P. is given by an = 3 + 4n. The common difference is

(a) 7

(b) 3

(c) 4

(d) 1

2. If p, q, r and s are in A.P. then r – q is

(a) s – p

(b) s – q

(c) s – r

(d) none of these

3. If the sum of three numbers in an A.P. is 9 and their product is 24, then numbers are

(a) 2, 4, 6

(b) 1, 5, 3

(c) 2, 8, 4

(d) 2, 3, 4

4. The (n – 1)th term of an A.P. is given by 7,12,17, 22,… is

(a) 5n + 2

(b) 5n + 3

(c) 5n – 5

(d) 5n – 3

5. The nth term of an A.P. 5, 2, -1, -4, -7 … is​

Answer 1

Answer:

1. c

2. c

3. d

4.d

5. Tn= (8-3n).

NOTE: Here Tn stands for "n th term of an AP".

Step-by-step explanation:

1. (c) Tn= 3+4n

therefore, T(n-1)= 3+4(n-1).

Now, d = Tn -T(n-1)= (3+4n)-[3+4(n-1)]=4.

2.(c) If p,q,r and s are in AP, then

q-p = r-q = s-r = common difference.

3.(d)Let the three nos. be (a-d),a,(a+d). Then,

(a-d)+a+(a+d)=9

3a = 9 or. a= 3.

and

(a-d).a.(a+d)= 24

or, a(a²-d²)=24

or, 3(3²-d²)= 24

or, 9-d²= 8

or, d²= 1

or, d= 1.

So, the AP is (3-1),3,(3+1), i.e., 2,3,4.

4.(d) Here, a= 7

d= 12-7=5.

therefore, (n-1) th term

= a +(n-1-1)d

= a+(n-2)d

=7+(n-2)×5

= 5n-3.

5. Here, a= 5 and d= (2-5)= -3.

So, nth term = a+(n-1)d

= 5+(n-1)×(-3)

= (8-3n).

Hence, Tn= 8-3n..