Question

In an $AP$ of $50$ term the sum of first $10$ terms is $210$ and the sum of its last $15$ terms is $2565$ Find the $AP$.

Answer 1

Sum of the first n terms is given by

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Putting n = 10, we get

$S_{10}=\frac{10}{2}[2a+(10-1\left)d\right]$

$210=5(2a+9d)$

$2a+9d=42$ ...............(1)

Sum of the last 15 terms is 2565

Sum of the first 50 terms - sum of the first 35 terms = 2565

$S_{50}-S_{35}=2565$

$\frac{50}{2}[2a+(50-1\left)d\right]-\frac{35}{2}[2a+(35-1\left)d\right]=2565$

$25(2a+49d)-\frac{35}{2}(2a+34d)=2565$

$5(2a+49d)-\frac{7}{2}(2a+34d)=513$

$10a+245d-7a+119d=513$

$3a+126d=513$

$a+42d=171$ ........(2)

Multiply the equation (2) with 2, we get

$2a+84d=342$ .........(3)

Subtracting (1) from (3)

$d=4$

Now, substituting the value of d in equation (1)

$2a+9d=42$

$2a+9\times 4=42$

$2a=42-36$

$2a=6$

$a=3$

So, the required AP is $3,7,11,15,19,23,27,31,35,39........$