Find the arithmetic progression whose third term is

Question:

Find the arithmetic progression whose third term is 16 and seventh term exceeds its fifth term by 12.

Solution:

Here, let us take the first term of the A.P as a and the common difference of the A.P as d

Now, as we know,

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

So, for $3^{\text {rd }}$ term $(n=3)$,

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

$16=a+2 d$

$a=16-2 d$.......(1)

Also, for 5th term (n = 5),

$a_{5}=a+(5-1) d$

$=a+4 d$

For 7th term (n = 7),

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

$=a+6 d$

Now, we are given,

$a_{7}=12+a_{5}$

$a+6 d=12+a+4 d$

$6 d-4 d=12$

$2 d=12$

$d=6$

Substituting the value of d in (1), we get,

$a=16-2(6)$

$=16-12$

$=4$

So, the first term is 4 and the common difference is 6.

Therefore, the A.P. is

 

 

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