Question

In an AP S5+S7=167 and S10=235 ,then find the S4​

Answer 1

Answer:

a + 4d + a + 6d = 167

2a + 10d = 167--- 1

a + 9d = 235 ---2

2 × 2

= 2a + 18d = 470---3

3-1

2a + 18d - 2a - 10d = 470-167

8d = 303

d = 37.875

from 2 : a + 9d=235

a + 9×37.875 = 235

a = -105.8

S4 = -105.8 + 3×37.875

S4 = 219.4

Answer 2

Step-by-step explanation:

${S}^{5} = 35$

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

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

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

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

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

${S}^{4} = 22$

$= > \frac{4}{2} [2a + (4 - 1)d] = 22$

$= > 2[2a + 3d] = 22$

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

On subtracting equation 1 and 2, we get

d = 3

Now

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

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

$= > 2a + 12 = 14$

$= > 2a = 2$

$= > a = 1$

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

$=1 + 4 \times 3$

$= 1 + 12$

$= 13$