Question

If the sum of first n terms of an AP is n square ,then find its 10th term.​

Answer 1

Answer:

a10=19

Step-by-step explanation:

sn=n^2

s1=a1

a1=1^2

a1=1

s2=a1+a2

2^2=1+a2

4-1=a2

3=a2

d=a2-a1

d=3-1

d=2

a10=a+9d

    =1+9(2)

   =1+18

    =19

Answer 2

Answer:  The required 10th term of the given A.P. is 19.

Step-by-step explanation:  Given that the sum of first n terms of an A.P. is n².

We are to find the 10th term of the A.P.

Let a be the first term and d be the common difference of the given A.P.

Then, he n-th term of the A.P. is given by

$a_n=a+(n-1)d$

and the sum of first n term of the A.P. is given by

$S_n=\dfrac{n}{2}(2a+(n-1)d).$

According to the given information, we have

$S_1=1^2\\\\\Rightarrow \dfrac{1}{2}(2a+(1-1)d)=1\\\\\\\Rightarrow \dfrac{1}{2}\times 2a=1\\\\\\\Rightarrow a=1$

and

$S_2=2^2\\\\\Rightarrow \dfrac{2}{2}(2a+(2-1)d)=4\\\\\\\Rightarrow2\times1+d=4\\\\\Rightarrow d=4-2\\\\\Rightarrow d=2.$

Therefore, the 10th term of the A.P. will be

$a_{10}=a+(10-1)d=1+9\times2=1+18=19.$

Thus, the required 10th term of the given A.P. is 19.