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

Question:

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.

Solution:

Given:

Let the first term of the A.P. be and the common difference be d.

$a_{4}=3 a$

$\Rightarrow a+(4-1) d=3 a$

$\Rightarrow a+3 d=3 a$

$\Rightarrow 3 d=2 a$

$\Rightarrow a=\frac{3 d}{2} \quad \ldots(\mathrm{i})$

And, $a_{7}-2 a_{3}=1$$\Rightarrow a+(7-1) d-2[a+(3-1) d]=1$

$\Rightarrow \mathrm{a}+6 d-2(a+2 d)=1$

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

$\Rightarrow-a+2 d=1$

$\Rightarrow-\frac{3 d}{2}+2 d=1 \quad[$ From (i) $]$

$\Rightarrow \frac{-3 d+4 d}{2}=1$

$\Rightarrow \frac{d}{2}=1$

$\Rightarrow d=2$

Putting the value in (i), we get:

$a=\frac{3 \times 2}{2}$

$\Rightarrow a=3$

 

 

 

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now