Question

If the nth term of a progression is (4n-10), show that it is an AP. Find is first term?

Answer 1

Given An......4n-10


Put n=1

4-10=-6

Put n=2
8-10=-2

Put n=3
12-10
2

Thus it forms an AP: -6,-2,2.....
Here a=-6 and d=4


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Answer 2

Answer:

The first term of AP is -6.

Step-by-step explanation:

The nth term of a progression is (4n-10). Thus, we have

$a_n=4n-10$

In order to check whether it an AP or not, we find some terms by substituting random values of n.

For n = 1

$a_1=4(1)-10\\\\a_1=-6$

For n =2

$a_2=4(2)-10\\\\a_2=-2$

For n =3

$a_3=4(3)-10\\\\a_3=2$

Thus, the sequence is

-10,-6,-2,2......

The difference of two consecutive terms should be equal in case of AP.

-2-(-6)=-2+6=4

2-(-2)=2+2=4

We can see the difference is equal. Hence, we can conclude that it is an AP.

The first term of AP is -6.