The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference.

Given:

The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1.

To do:

We have to find the first term and the common difference.

Solution:

Let the first term, common difference and the number of terms of the given A.P. be $a, d$ and $n$ respectively.

We know that,

nth term of an A.P. $a_n=a+(n-1)d$

Therefore,

First term $a_1=a$

$a_{4}=a+(4-1)d$

$=a+3d$.....(i)

$a_{7}=a+(7-1)d$

$=a+6d$....(ii)

$a_{3}=a+(3-1)d$

$=a+2d$....(iii)

According to the question,

$a_4=3\times a_1$

$a+3d=3a$

$3a-a=3d$

$2a=3d$

$a=\frac{3d}{2}$....(iv)

$a_7=2\times a_3+1$

$a+6d=2(a+2d)+1$

$a+6d=2a+4d+1$

$2a-a+1=6d-4d$

$a+1=2d$

$\frac{3d}{2}+1=2d$

$\frac{3d+2}{2}=2d$

$3d+2=2(2d)$

$3d+2=4d$

$4d-3d=2$

$d=2$....(v)

Substituting $d=2$ in (iv), we get,

$a=\frac{3(2)}{2}$

$a=3$

The first term is $3$ and the common difference is $2$.

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Updated on: 10-Oct-2022

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