# A1,a2,a3,...,a40 are the terms of am AP and a1+a5+a15+a26+a36+a40=105 then find a1+a2+a3+....+a40 ?

Aug 1, 2018

$700$.

#### Explanation:

Suppose that the common difference of the AP in question is $d$.

$\therefore {a}_{n} = {a}_{1} + \left(n - 1\right) d , n \in \mathbb{N}$.

Given that, ${a}_{1} + {a}_{5} + {a}_{15} + {a}_{26} + {a}_{36} + {a}_{40} = 105 ,$

$\Rightarrow {a}_{1} + \left\{{a}_{1} + 4 d\right\} + \left\{{a}_{1} + 14 d\right\} + \left\{{a}_{1} + 25 d\right\} + \left\{{a}_{1} + 35 d\right\} + \left\{{a}_{1} + 39 d\right\} = 105$.

$\therefore 6 {a}_{1} + 117 d = 105$.

Dividing by $3 , 2 {a}_{1} + 39 d = 35. \ldots \ldots \ldots \ldots \ldots \left(\star\right)$.

Recall that, ${a}_{1} + {a}_{2} + \ldots + {a}_{n} = \frac{n}{2} \left\{2 {a}_{1} + \left(n - 1\right) d\right\}$.

$\therefore {a}_{1} + {a}_{2} + \ldots + {a}_{40} = \frac{40}{2} \left\{2 {a}_{1} + 39 d\right\}$,

$= 20 \left(35\right) \ldots \ldots \ldots \ldots \ldots \left[\because , \left(\star\right)\right]$,

$= 700$.