Prove that

Question:

The $4^{\text {th }}$ term of an AP is three times the first and the $7^{\text {th }}$ term exceeds twice the third term by 1. Find the first term and the common difference.

 

Solution:

To Find: First term (a) and common difference (d)

Given: $a_{4}=3 a_{1}$ and $a_{7}=2 a_{3}+1$

(Where $a=a_{1}$ is first term, $a_{n}$ is nth term and $d$ is common difference of given $A P$ )

Formula Used: $a_{n}=a+(n-1) d$

$a_{4}=3 a_{1} \rightarrow a+3 d=3 a \rightarrow 3 d=2 a \ldots$ equation (i) and

$a_{7}=2 a_{3}+1 \rightarrow a+6 d=2(a+2 d)+1 \rightarrow 2 d=a+1 \ldots$ equation (ii)

On solving both equation (i) & (ii), we get

$a=3$ and $d=2$

So the first term is equal to 3, and the common difference is equal to 2.

 

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