Question

If in an A.P S5=35 & S4=22FIND THE 5TH TERM

Answer 1

Since $S_{5}=35$

Sum of 'n' terms of an AP is given by the formula $S_{n}=\frac{n}{2}[2a+(n-1)d]$

Now, $S_{5}=\frac{5}{2}[2a+(5-1)d]=35$

$[2a+(5-1)d]=\frac{35 \times 2}{5}$

$2a+4d=14$

a+2d=7

a = 7 - 2d(Equation 1)

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

2a+3d = 11    (Equation 2)

Substituting the value of 'a'  from equation 1 in equation 2.

2(7-2d) + 3d =11

14 - 4d + 3d =11

14 -d = 11

14 - 11 = d

d = 3

Now, a = 7 - 2d

a = 7- (2 x 3)

= 7 - 6 = 1

Now, we have to determine the fifth term of an AP,

$a_{n}=a+(n-1)d$

$a_{5}=a+(5-1)d= a+ 4d$

= 1+(4 x 3)

= 13

So, the fifth term of an AP is 13.

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

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

$= > 2a + 12 = 14$

$= > 2a = 2$

$= > a = 1$

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

$= 3 + 4 \times 2$

$= 3 + 8$

$= 11$