Question

in an AP. if S5 is 35 and S4 is 22 then the fifth term of AP is

Answer 1

let 'a' be the first term and 'd' be the common difference

S5 = a+(5-1)d = 35
=a+4d = 35 -------(1)
also ,
S4 = a+(4-1)d= 22
=a+3d= 22-------(2)
subtracting (1) and (2) we get

a+4d=35
a+3d=22
-----------------
d= 13


put d= 13 in equation 1

a+4(13)= 35
a+52=35
a=35-52
a= -17

let's find fifth term of an ap
by formula,

an= a+(n-1)d

= -17+(5-1)13

= -17+(4)13

= -17+ 52

= 35

fifth term of an ap =35

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

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

$= > 2a + 12 = 14$

$= > 2a = 2$

$= > a = 1$

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

$= 3 + 4 \times 2$

$= 3 + 8$

$= 11$