Question

\[5, 8, 11, 14, .....\] are in A.P.

Assertion (A): \[\frac{5}{2}, 4, \frac{11}{2}, 7, .....\] are also in A.P.

Reason (R): If each term of a given A.P. is divided by the same non-zero number, the resulting sequence is an A.P.

विकल्प

• A is true, R is false.

• A is false, R is true.

• Both A and R are true and R is the correct reason for A.

• Both A and R are true and R is the incorrect reason for A.

Answer 1

Both A and R are true and R is the correct reason for A.

Explanation:

Given, 5, 8, 11, 14, ............... are in AP.

Here, first term \[{} = 5\], common difference 8 − 5 = 11 − 8 = 3

Now, new sequence : \[\frac{5}{2}, 4, \frac{11}{2}, 7, .................\]

The above sequence is found by dividing 5, 8, 11, 14, .............. the sequence by 2.

In new sequence,

Here,

Difference between second and first term \[{} = 4 - \frac{5}{2} = \frac{8 - 5}{2} = \frac{3}{2}\]

Difference between third and second term \[{} = \frac{11}{2} - 4 = \frac{11 - 8}{2} = \frac{3}{2}\]

Difference between fourth and third term \[{} = 7 - \frac{11}{2} = \frac{14 - 11}{2} = \frac{3}{2}\]

So, the common difference is same, means the given sequence is also in A.P..

So, Assertion is true.

The sequence \[\frac{5}{2}, 4, \frac{11}{2}, 7, .................\] this sequence is found by each term of the A.P. 5, 8, 11, 14, ............... is divided by 2.

If each term of a given A.P. is divided by the same non-zero number, the resulting sequence is an A.P.

So, Reason is true.