Question

in an AP if S5=35 and S4=22 then the 5th term is

Answer 1

S5 = 35

=> n/2 [ 2a +(n-1)d] = 35

=> 5/2 [ 2a + 4d] = 35

=> 2a +4d = 35 × 2 / 5

=> 2a + 4d = 7 × 2

=> 2a + 4d = 14 -------(1)


S4 = 24


=> 4/2 [ 2a + (4-1)d] = 24

=> 2 [ 2a + 3d] = 24

=> 2a + 3d = 12 ---------(2)

On subtracting equation 1 and 2, we get

d = 2

Now,

On Substituting the value of d in equation 1, we get

2a + 4× 2 = 14

=> 2a + 8 = 14

=> 2a = 6

=> a = 3

T5 = a + 4d

= 3 + 4 × 2

= 3 + 8

= 11

Answer 2

Answer:

${S}^{5} = 35$

$= > \frac{n}{2} [2a + (n - 1)d] = 35$

$= > \frac{5}{2} [2a + 4d] = 35$

$= > 2a + 4d = 35 \times \frac{2}{5}$

$= > 2a + 4d = 7 \times 2$

$= > 2a + 4d = 14 \: \: \: \: \: \: \: \: \: \: --------1$

${S}^{4} = 24$

$= > \frac{4}{2} [2a + (4 - 1)d] = 24$

$= > 2[2a + 3d] = 24$

$= > 2a + 3d = 11 \: \: \: \: \: \: \: \: \: \: -------- \: 2$

On subtracting equation 1 and 2, we get

d = 3

Now

On substituting the value of d in equation 1, we get

$= > 2a + 4 \times 3 = 14$

$= > 2a + 12 = 14$

$= > 2a = 2$

$= > a = 1$

$5th\:term = a + 4d$

$= 3 + 4 \times 2$

$= 3 + 8$

$= 11$