# What is the common difference? if an A. P has the sum of first and last term is 43 and the third is 11 the sum is 258

Jun 1, 2018

The first term is $\frac{35}{18}$

The common difference is $\frac{32}{9}$

#### Explanation:

Suppose we denote the ${n}^{t h}$ term of the AP by ${u}_{n}$ where ${u}_{n} = a + n d$, making the AP sequence:

$\left\{a , a + d , a + 2 d , a + 3 d , \ldots , a + \left(n - 1\right) d\right\}$

Given the sum of the first and last term is $43$, we have:

$\left(a\right) + \left(a + \left(n - 1\right) d\right) = 43$

$\therefore 2 a + \left(n - 1\right) d = 43$ ..... [A]

Given, also, that the third term is $11$, we have:

$a + 2 d = 11$ ..... [B]

Given, also, that the sum is $258$, and utilising the AP summation formula:

${S}_{n} = \frac{n}{2} \left(2 a + \left(n - 1\right) d\right)$

We have:

$\frac{n}{2} \left(2 a + \left(n - 1\right) d\right) = 258$

$n \left(2 a + \left(n - 1\right) d\right) = 516$ ..... [C]

Substituting [A] into [C] we get:

$n \left(43\right) = 516 \implies n = 12$

Back substitution of $n = 12$ into [A] we get:

$\therefore 2 a + 11 d = 43$ ..... [D]

So that [D]-2[B] yields:

$11 d - 2 d = 43 - 11 \implies 9 d = 32$

$\therefore d = \frac{32}{9}$

Finally, substituting $d = \frac{32}{9}$ into [D]:

$2 a + \frac{11 \cdot 32}{9} = 43 \implies 2 a = 43 - \frac{352}{9}$

$\therefore a = \frac{35}{18}$