Question

Sum of first 5 terms of an ap is 25,sum of next 5terms is -75 .Find 10th term of ap

Answer 1

Answer:

sum = n/2 [ 2a + (n-1)d]= 25

n (number of terms)= 5

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

5/2 [ 2a + (4)d]= 25

[ 2a + (4)d]= (25*2)/ 5

[ 2a + (4)d]=10

2a + 4d= 10................(i)

sum of next 5 terms = -75

n=10

n/2 [ 2a + (n-1)d] - [2a + 4d= 10]= -75

10/2 [ 2a + (10-1)d] - [2a + 4d= 10]= -75

5 [ 2a + (9)d] - [2a + 4d= 10]= -75

[10a+45d] - 10 = -75

[10a+45d]= -85................. (ii)

dividing equation (ii) by 5

[2a+9d]= -17.............(iii)

∴(i)-(ii)

-5d= 27

d= -27/5

a=7.7

10th term= 7.7+9*-27/5

= 56.3(ans)

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Let S

n

be the sum of n terms of an AP with first term a and common difference d. Then

S

n

=

2

n

(2a+(n−1)d)

putting n=5 in S

n

, we get

S

5

=

2

5

(2a+4d)⇒S

5

=5(a+2d) ...(1)

putting n=7 in S

n

, we get

S

7

=

2

7

(2a+6d)⇒S

7

=7(a+3d) ...(2)

Adding (1) and (2), we get

12a+31d=167 ...(3)

Now,S

10

=235⇒

2

10

(2a+9d)=235

⇒2a+9d=47 ...(4)

On applying 6×(4)−(3), we get

23d=115⇒d=5

Putting d=5 in (3), we get

12a=167−31×5⇒a=1

Therefore, Required sum S

20

=

2

20

[2×1+(20−1)×5

⇒S

20

=10[2+95]

⇒S

20

=970

Hope its help..