 # Arithmetic Progression

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• Q1

If a, b, c are in A.P., then prove that ka, kb, kc are in A.P.

Marks:2

If a, b, c are in A.P. then

b-a = c-b ...(i)

Now, ka, kb and kc will be in A.P.

If kb- ka = kc- kb

or k(b-a) = k(c-b)

or b-a = c-b which is true by (i) .

Hence if a, b, c are in A.P., then ka, kb, kc are also in A.P.

• Q2

If 20th term of an A.P. is 99 and 3rd term is 14, find its first and 4th term.

Marks:2

Tn = a + (n - 1)d

T
20 = 99 = a + (20 - 1)d
a + 19d= 99 ...(i)

T3 = 14 = a + (3 - 1)d
a + 2d = 14 ...(ii)

Solving these two equations we get, a = 4 and d = 5.

Thus the first term is 4 and T4 = 4 + (4 - 1)5
= 19.

• Q3

Find the 12th term of an A.P. whose first term is 2 and common difference is 4.

Marks:1

For an A.P.

Tn = a + (n - 1)d

Here a = 2, d = 4 and n = 12.

T12 = 2 + (12 - 1)4 = 2 + 44 = 46.

• Q4

The nth term of a sequence is given by an=n2+3n-5. Find its 21st term.

Marks:2

We have,

an=n2+3n-5

On putting n=21, we get

21st term=212+3x21-5

= 441+63-5

= 504-5

= 499  