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Q.

The fourth term of an A.P is three times of the first term and the seventh term exceeds the twice of the third term by one, then the common difference of the progression is

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a

$\frac{3}{2}$

b

2

c

3

d

–1

answer is A.

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Detailed Solution

Let the progression be $a, a + d, a + 2 d . . . .$

$g i v e n t h a t t_{4} = 3 t_{1} \Rightarrow a + 3 d = 3 a \Rightarrow 3 d = 2 a \to (i)$

Again $t_{7} = 2 t_{3} + 1$

$\Rightarrow a + 6 d = 2 (a + 2 d) + 1 \Rightarrow 2 d = a + 1 \to (i i)$

Solving (i) and (ii), then we get a= 3, d= 2

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The fourth term of an A.P is three times of the first term and the seventh term exceeds the twice of the third term by one, then the common difference of the progression is