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Sum of n Terms

According to ...

Question

According to the property of A.P. If each term of a given A.P $(1, 2, 3, . . .)$ is divided by a non-zero constant $t$ , then the resulting sequence will be

A.P

(\frac{1}{t}, \frac{2}{t}, \frac{3}{t} \dots)

, with common difference

\frac{1}{t}

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G.P

(\frac{1}{t}, \frac{2}{t}, \frac{3}{t} \dots)

, with common difference

\frac{1}{t}

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H.P

(\frac{1}{t}, \frac{2}{t}, \frac{3}{t} \dots)

, with common difference

\frac{1}{t}

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A.P

(t, 2 t, 3 t \dots)

, with common difference

t

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Solution

The correct option is B A.P $(\frac{1}{t}, \frac{2}{t}, \frac{3}{t} \dots)$ , with common difference $\frac{1}{t}$
Given AP is $1, 2, 3, 4...$
If each term is divided by $t$ , then the series will be:
$\frac{1}{t}, \frac{2}{t}, \frac{3}{t} . . .$
The new series is also an AP with common difference $d = \frac{2}{t} - \frac{1}{t} = \frac{1}{t}$
Hence, option A is correct.

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According to the property of A.P. If each term of a given A.P (1,2,3,...) is divided by a non-zero constant t, then the resulting sequence will be

The correct option is B A.P (1t,2t,3t…), with common difference1tGiven AP is 1,2,3,4...If each term is divided by t, then the series will be:1t,2t,3t...The new series is also an AP with common difference d=2t−1t=1tHence, option A is correct.

According to the property of A.P. If each term of a given A.P $(1, 2, 3, . . .)$ is divided by a non-zero constant $t$ , then the resulting sequence will be