Question

In A.P Sum of first 10 consecutive is 25and the common difference is twice of first term then find the 10th term

Answer 1

sn= n/2( a+(n-1)d)
s10=5(a+(9)2a
25=5(19a)
a=5/19

d=10/19
T10=5/19+(9)10/19

Answer 2

Heya,

According to the question,

In A.P. sum of first 10 consecutive no. is 25
S10 = 10/2(2a + (10 - 1)d). .... equation 1

S10 = 25

And,
common difference is twice the first term.
d = 2a

Put value of d in equation 1

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

=> S10 = 5(2a + 9×2a)

=> S10 = 5(2a +18a)

=> S10 = 5×20a

=> 25 = 100a

=> a = 25/100

=> a = 1/4

Therefore,

d = 2(1/4)

d = 1/2

Now,

Tenth term of A.P., will be,

An = a + (n - 1)d

=> A10 = 1/4 + (10 - 1)1/2

=> A10 = 1/4 + 9/2

=> A10 = (1 + 18)/4

=> A10 = 19/4. ..... Answer

Hope this helps....:)