Question

can the sum of nth terms in an AP is n2 +13n justify your answer​

Answer 1

Answer:

no

n²+13n

put the value in n =,2,3,..

1 case 14

2 case 30

3case48

because common difference is not same

Answer 2

Yes, the sum of nth term in an AP is $n^2+13n$

Step-by-step explanation:

Since we have given that

$S_n=n^2+13n$

Now, we put n = 1, we get that

$S_1=(1)^2+13(1)=1+13=14=a$ ( first term)

now, we put n = 2, we get

$S_2=(2)^2+13\times 2\\\\S_2=4+26=30=a+a_2$

So, we get

$30=14+a_2\\\\30-14=a_2\\\\16=a_2$

We put n = 3, we get

$S_3=(3)^2+13(3)=9+39=48=a+a_2+a_3$

so, it becomes,

$48=14+16+a_3\\\\48=30+a_3\\\\48-30=a_3\\\\18=a_3$

Since there is common difference of 2 in each term, so it can form A.P.

Hence, Yes, the sum of nth term in an AP is $n^2+13n$

# learn more:

The nth term of an AP cannot be n2 + 1. Justify your answer

https://brainly.in/question/278253