# Determine the first and last terms of an arithmetic series with 50 terms, a common difference of 6, and a sum of 7850. Help?

$10$ & $314$

#### Explanation:

Let $a$ & $l$ be the first & last terms respectively of an A.P.

Now, the last 50th term of series having $n = 50$ terms with a common difference $d = 6$ will be $l$

$\setminus \therefore l = a + \left(n - 1\right) d$

$l = a + \left(50 - 1\right) 6$
$l - a = 294 \setminus \ldots \ldots . \left(1\right)$

Now, the sum of AP with $n = 50$ terms is 7850

$\setminus \therefore 7850 = \setminus \frac{n}{2} \left(a + l\right)$

$7850 = \setminus \frac{50}{2} \left(a + l\right)$
$a + l = 314 \setminus \ldots \ldots . \left(2\right)$

Adding (1) & (2), we get

$2 l = 608$

$l = 304$ &

$a = 314 - l = 314 - 304 = 10$

hence, the first & last terms of given AP. are $10$ & $304$

Jun 24, 2018

First term =10 and Last term =304

#### Explanation:

We have No. of terms=50;C.D=6;${S}_{n}$=7850

using ${S}_{n} = n \cdot \frac{2 \cdot a + \left(n - 1\right) \cdot d}{2}$
now putting the values

$50 \cdot \frac{2 \cdot a + \left(50 - 1\right) 6}{2} = 7850$

$50 \cdot a + 7350 = 7850$
$50 a = 500$
$a = 10$
now again using ${S}_{n} = n \cdot \frac{a + l}{2}$ where l=last term

$25 \cdot \left(10 + l\right) = 7850$
$10 + l = 314$
l=last term =$304$