Question

If s5 and s4 of an AP are 25 and 16 respectively find a5

Answer 1

Answer:

sn = n/2 (2a + (n-1) d)

S5 =25

25 =5/2 (2a +(5-1) d)

25 ×2 = 5 (2a + 4d)

50/5 = 2a +4d

10= 2a + 4d eq 1

S4 = 16

16 = 4/2 (2a +(4-1) d)

16 = 2 (2a + 3d)

16/2 =2a + 3d

8 = 2a + 3d eq 2

subtract eq 1 and 2

10 = 2a + 4d

8 = 2a + 3d

2 = d

put the value of d in equation 2

8 = 2a + 3d

8 = 2a +3(2)

8 = 2a + 6

8-6 = 2a

2/2 = a

1 = a

a5 = a+4d

a5 = 1+4(2)

a5 = 1+ 8

a5= 9

hope it helps you

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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

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

$= > 2a + 12 = 14$

$= > 2a = 2$

$= > a = 1$

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

$= 3 + 4 \times 2$

$= 3 + 8$

$= 11$