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In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the A.P.
Given: An A.P. of 50 terms, and sum of its first 10 terms$=210$ and sum of its last 15 terms $=2565$.
To do: To find the A.P.
Solution: Let a and d be the first term and the common difference of an A. P. respectively.
n term of an A.P., $a_{n}=a+(n-1)d$
Sum of n terms of an A.P., $S_{n}=\frac{n}{2}[ 2a+( n-1) d]$
We have Sum of the first 10 terms $S_{10} =\frac{10}{2}[ 2a+( 10-1) d]$
$S_{10} =5( 2a+9d)$
$\Rightarrow 10a+45d=210$
$\Rightarrow 2a+9d=42\ ................( 1)$
And its $35^{th}$ term, $a_{35} =a+( 35-1) d$
$a_{35} =a+34d$
Similarly ,its $50^{th}$ term $a_{50} =a+49d$
Here for the last 15 term, $a_{35}$ is the first term and $a_{50}$ is the last term.
common difference will be the same.
sum of the last 15 terms$=\frac{n}{2}( a+l) \ \ \ \ \ \ \ \ \ \ \ ( l\ stands\ for\ last\ term\ of\ the\ A.P.)$
$\Rightarrow \frac{15}{2}( a+35d+a+49d) =2565$
$\Rightarrow 2a+84d=171\times 2$
$\Rightarrow 2a+84d=342$
$a+42d=171\ \ \ \ .....................( 2)$
On solving $( 1)$ and $( 2)$ ,
We have $a=3$ and $d=4$
Therefore the A.P. $3,\ 7,\ 11,\ 15,............199
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