The 4th term of an A.P. is equal to 3 times t | vedclass

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(A) Let $a$ be the first term and $d$ be the common difference of the $A.P.$The $4$th term is $a_4 = a + 3d$. According to the problem,$a_4 = 3a$.So,$

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$3, 2$

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(A) Let $a$ be the first term and $d$ be the common difference of the $A.P.$The $4$th term is $a_4 = a + 3d$. According to the problem,$a_4 = 3a$.So,$a + 3d = 3a \Rightarrow 3d = 2a \Rightarrow a = \frac{3}{2}d$ $...(1)$The $7$th term is $a_7 = a + 6d$ and the $3$rd term is $a_3 = a + 2d$.According to the problem,$a_7 = 2a_3 + 1$.So,$a + 6d = 2(a + 2d) + 1$$a + 6d = 2a + 4d + 1$$2d - a = 1$ $...(2)$Substitute $a = \frac{3}{2}d$ from equation $(1)$ into equation $(2)$:$2d - \frac{3}{2}d = 1$$\frac{4d - 3d}{2} = 1$$\frac{d}{2} = 1 \Rightarrow d = 2$Now,substitute $d = 2$ into equation $(1)$:$a = \frac{3}{2} 	imes 2 = 3$Thus,the first term is $3$ and the common difference is $2$.

The $4$th term of an $A.P.$ is equal to $3$ times the first term and the seventh term exceeds twice the third term by $1$. Find the first term and the common difference.

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