Question

In an AP of 50 terms the sum of the first 10 terms is 210 and the sum of the last 15 terms is 2565. find the AP.

Answer 1

Consider a and d as the first term and the common difference of an A.P. respectively.

$n^{th}$ term of an A.P., an = a + ( n – 1)d

Sum of n terms of an A.P., $S_n=\frac{n}{2}[2a+(n–1\left)d\right]$

Given that the sum of the first 10 terms is 210.

$\Rightarrow \frac{10}{2}[2a+9d]=210$

$\Rightarrow 5[2a+9d]=210$

$\Rightarrow 2a+9d=42\dots \dots \left(i\right)$

${15}^{th}$ term from the last $=(50–15+1{)}^{th}={36}^{th}$ term from the beginning

$\Rightarrow a_{36}=a+35d$

Sum of the last $15$ terms $=\frac{15}{2}[2a_{36}+(15–1\left)d\right]=2565$ [Starting from $a_{36}$]

$\Rightarrow \frac{15}{2}\left[2\right(a+35d)+14d]=2565$

$\Rightarrow 15[a+35d+7d]=2565$

$\Rightarrow a+42d=171\dots \dots \left(ii\right)$

On multiplying eq.(ii) by 2 and subtracting the result from eq.(i), we get

$\Rightarrow 2a+9d-2(a+42d)=42-2\left(171\right)$

$\Rightarrow 9d-84d=42-342$

$\Rightarrow -75d=-300$

$\Rightarrow d=4$

And putting the value of d in eq.(i), we get $a=3.$

Therefore, the terms in A.P. are $3,7,11,15...199.$