Question

An $AP$ consists of $21$ terms. The sum of the three terms in the middle is $129$ & of the last three is $237$. Find $AP$.

Answer 1

Total terms $=21$

Middle term $={11}^{th}$ term

So, three middle terms are ${10}^{th},{11}^{th},{12}^{th}$ term.

Therefore,

$a_{10}+a_{11}+a_{12}=129$

$a+9d+a+10d+a+11d=129$

$a+10d=43$ ---- (i)

also,

$a_{19}+a_{20}+a_{21}=237$

Similarly,

$a+19d=79$ ----- (ii)

subtracting (i) from (ii), we get

$d=4$

From (i)

$a+10d=43$

$a=3$

So, A.P. is $3,7,11,15,.....$