Question

If S5=25 and S4=20 findA5

​

Answer 1

Answer:

Let the first term is a and the common difference is d

By using S

n

=

2

n

[2a+(n−1)d] we have,

S

5

=

2

5

[2a+(5−1)d]=

2

5

[2a+4d]

S

7

=

2

7

[2a+(7−1)d]=

2

7

[2a+6d]

Given: S

7

+S

5

=167

∴

2

5

[2a+4d]+

2

7

[2a+6d]=167

⇒10a+20d+14a+42d=334

⇒24a+62d=334 ...(1)

S

10

=

2

10

[2a+(10−1)d]=5(2a+9d)

Given: S

10

=235

So 5(2a+9d)=235

⇒2a+9d=47 ...(2)

Multiply equation (2) by 12, we get

24a+108d=564....(3)

Subtracting equation (3) from (1), we get

−46d=−230

∴d=5

Substing the value of d=5 in equation (1) we get

2a+9(5)=47 or 2a=2

∴a=1

Then A.P is 1,6,11,16,21,

Answer 2

Answer:

Let the first term of the A.P. is a and the

common difference is d

Given that, the sum of first 5 terms

i.e.,5/2^ * \ 2a+(5-1)d\ =30

or,5/2^ * (2a+4d)=30

or, a + 2d = 6 (1)

Again the sum of the first 4 terms is S_{4} = 20

i.e.,4/2^ * \ 2a+(4-1)d\ =20

or, 2(2a + 3d) = 20

or, 2a + 3d = 10

or, 2(6 - 2d) + 3d = 10, by * (1)

or, 12 - 4d + 3d = 10

or, d = 2

From (1), we write

a + 2(2) = 6

or, a + 4 = 6

or, a = 2

Hence the 5th term of the A.P. is

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

=2+4(2)

=2+8

о

=10

Step-by-step explanation:

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