Question

If an $A.P.$ has $21$ terms , and the sum of ${10}^{th},{11}^{th}$ & ${12}^{th}$ terms is $129$ and sum of last three terms is $237$. Find first term.

Answer 1

$n^{th}$ term $=t_n=a+(n-1)d$

$a=$ First term

$d=$ Common difference

$n=$ number of terms

$t_n=n^{th}$ term

Total terms $=21$

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

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

$\Rightarrow 3a+30d=129$\

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

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

$\Rightarrow a+20d+a+19d+a+18d=237$

$\Rightarrow 3a+57d=237$

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

On subtracting (i) from (ii) we get

$d=4$

putting $d=4$ in (i) we get

$a+10\times 4=43$

$\Rightarrow a=3$

Therefore, first term of the given A.P is $3$