Question

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.

Answer 1

Let a and b the first term and the common difference of an A.P. respectively.
$n^{th}$ term of an A.P, $a_n=a(n-1)d$
Sum of n terms of an A.P, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
We have :
Sum of the first 10 terms $\frac{10}{2}[2a+9d]$
$\Rightarrow 210=5[2a+9d]$

⇒ 42 = 2a+9d ………. (1)

15th term from the last = ( 50−15 + 1) th= 36th term from the beginning

Now, $a_{36}=a+35d$

∴ Sum of the last 15 terms =$\frac{15}{2}$ ($2a_{36}+(15-1)d$)

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

=15[a +35d +7d]

⇒2565 = 15[a+42d]

⇒ 171= a+42d ………….(2)

From (1) and (2), we get,

d = 4

a = 3

So, the A. P. formed is 3, 7, 11, 15 …. and 199.