Question

In an A.P.,S4=35 AND S4 =22 find the 5th term

Answer 1

since sn = n/2{2a+(n-1)d}
s4=4/2{2a + (4-1)d}
35 = 4/2 × {2a + 3d}
70 = 4 × {2a + 3d}
70 = 8a + 12d
35 = 4a + 6d ..........( dividing by 2 on both side)
4a + 6d = 35
a + 6d = 7 ............(1)
s4 = 4/2{2a + (4-1)d}
22 = 4/2 {2a + 3d}
44= 8a + 12d
22 = 4a + 6d
4a + 6d = 22...........(2)


subtracting eq. 2 from 1

a + 6d= 7
4a+6d=22
(-). (-). (-)
_________
-3a. = -15
a. = 15/3
a = 5


substitute a = 5 in eq. 1

a + 6d = 7
5 + 6d = 7
6d = 2
d = 2/6
d= 1/3

t5 = a + (n-1)d
t5 = 5 + 5-1 × 1/3
= 5 + 4 × 1/3
= 9 × 1/3
t5 = 3

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

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

$= > 2a + 12 = 14$

$= > 2a = 2$

$= > a = 1$

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

$= 1+ 4 \times 3$

$= 1 + 12$

$= 13$