Question

Determine the number of terms in tha A.P 3 7 11 ...407 also find 20th term of The end

Answer 1

3,7,11,.................,407

407= 3+(n-1)(4)    nth term = a+(n-1)d

 = 3+4n-4

 = 4n-1

 408=4n

n=408/4

n=102

Number of terms = 102

It we start from last

Then first term =407

common difference =(-4)

20th term = 407 + (20-1) (-4)

 = 407+ 19 * (-4)

 = 407 - 76  

     = 331 ans.

Answer 2

Answer:

Required number of terms are 102 and 20th term of the end is 83.

Step-by-step explanation:

Given Series is 3,7,11,....,407.

Here first term of series is 3 and common difference is (7-3)=4 and last term is 407.

We know,

$T_n = a + (n - 1)d.....(1)$

Where,$T_n$=Last term,a=First term and d is the common difference and n is the number of terms.

Here,

$T_n = 407$

$a = 3 \\ d = 4$

So,

$407 = 3 + (n - 1)4 \\ 407 = 3 + 4n - 4 \\ 4n = 408 \\ n = \frac{408}{4} \\ n = 102$

Therefore total number of terms 102.

We need to find 20 th term from the end.

Last term = 102th

Second last term = 102-1 =101 th

Third last term = 102-2 = 100

So,we can say pth last term = 102-(p-1)

So,20th term from the end

$= 102 - (20 - 1) = 102 - 19 = 83$