Question

The sum of the first 10 terms of an AP is 25 and the common difference is twice the first term. find the 10th term

Answer 1

Answer: 11/2 or 5.5

Step-by-step explanation:

let the first term of the AP be a and the common difference be d

d=2a (given) -----1

S10= 25 (given)

10 (2a+(10-1)d)=25  (∵sum of n terms= n/2 (2a+(n-1)d))

2

5(d+9d)=25  (by 1)

5d+45d=25

50d=25

d= 1

    2

2x 1/2= a (by 1)

a=1

A10= a+(10-1)d   (∵An= a+(n-1)d)

     = 1+9(1/2)

     =1 + 9/2

     = 11/2

     = 5.5

Answer 2

Answer: 19/4

Step-by-step explanation:

Given,

S10 = n/2(2a+(10-1)d) -----------(1)

S10= 25

d= 2a

n = 10

Putting the value of 'd' in equation (1):

25= 10/2 (2a+(10-1)2a)

25= 10/2 (2a+9×2a)

25= 10/2 (2a+18a)

25= 10/2 (20a)

25= 5×20a

25= 100a

25/100=a

a= 1/4

Now,

d= 2a

d= 2×1/4

d= 1/2

Now, the 10th term of the A.P. will be:

an = a+(n-1)d

a10= 1/4+(10-1)1/2

a10= 1/4 +9×1/2

a10= 1/4+9/2

a10= 19/4