Question

given last term 4, d= 2 sum of n term 14 find N and A​

Answer 1

Answer:

$4 = a + (n - 1) d$

$a + (n - 1) \times 2 = 4$

$a + 2n = 6.........(i)$

now,

$s_{n} = 14$

$\frac{n}{2} \times (a + l) = 14$

$\frac{n}{2} \times (a + 4) = 14$

putting the value of a from (i), a = 6–2n

$n \times (6 - 2n + 4) = 28$

$10n - 2 {n}^{2} = 28$

$2 {n}^{2} - 10n + 28 = 0$

${n}^{2} - 5n + 14 = 0$

it hasn't give real value of n so, not possible.

hence we can't find real a & n.