Question

i) The seventh term of an A.P. is 30 and the tenth term is 21, find the fourth term.
ii) The third term of an A.P. is -11 and the nineth term is -35, find the nth term of the A.P.

Answer 1

answer for 1st question hope will help you

Answer 2

Answer:-

(i)Fourth term of the A.P is 39

(ii)Nth term of the A.P is $a_{n}\ =\ -3\ +\ (n-1)(-4)$

Step-by-step explanation:

(i)In this question,

We have been given that seventh term of the A.P is 30 and Tenth term is 21.We need to find the Fourth term

Formula for nth term of the A.P is

$a_{n}\ =\ a\ +\ (n-1)d$

Here,

$a_{7}\ =\ 30\ n\ =\ 7$

Therefore $a_{7}\ =\ a\ +\ (7-1)d$

                         30 = a + 6d  .......(A)

Similarly, $a_{10}\ =\ 21\ n\ =\ 10$

Therefore $a_{10}\ =\ a\ +\ (10-1)d$

                            21 = a + 9d ........(B)

On subtracting B - A we get,

-9 = 3d

Therefore d = -3

Putting the value of d in (A) we get,

30 = a + 6(-3)

30 = a - 18

Therefore a = 48

We need to find the 4th term

$a_{4}\ =\ 48\ +\ (4-1)(-3)$

$a_{4}\ =\ 48\ +\ 3(-3)$

$a_{4}\ =\ 48\ -\ 9$

$a_{4}\ =\ 39$

Hence fourth term is 39

(ii) In this question

We have been given that 3rd term of the A.P is -11 and nineth term is -35

Then we need to find the Nth term of the A.P

Formula for the nth term of the A.P

$a_{n}\ =\ a\ +\ (n-1)d$

Here,

$a_{3}\ =\ -11\ n\ =\ 3$

Therefore $a_{3}\ =\ a\ +\ (3-1)d$

                        -11 = a + 2d  .......(A)

Similarly, $a_{9}\ =\ -35\ n\ =\ 9$

Therefore $a_{0}\ =\ a\ +\ (9-1)d$

                            -35 = a + 8d ........(B)

On subtracting (A)-(B) we get,

24 = -6d

Therefore d = -4

Putting the value of d in (A) we get

-11 = a -8

Therefore a = -3

Nth term of the A.P is

$a_{n}\ =\ -3\ +\ (n-1)(-4)$