Question

The 5th term and 21st term of a AP are 10 and 42 respectively find the 31st term

Answer 1

$\large{ \bold{ \underline { \underline{ \: Answer : \: \: \: }}}}$

$\to$ 31st term = 62

$\large{ \bold{ \underline { \underline{ \: To \: Find : \: \: \: }}}}$

$\to$ 31st term = ?

$\large{ \bold{ \underline { \underline{ \: Solution : \: \: \: \: }}}}$

Let ,

$\to$ First term = a

$\to$ Common difference = d

A.T.Q

$\to$ 5th term = 10

$\to$ a + ( 5 - 1 )d = 10

$\to$ a + 4d = 10 ------- eq (i)

$\to$ 21st term = 42

$\to$ a + ( 21 - 1 )d = 42

$\to$ a + 20d = 42 ------- eq (ii)

Equation (ii) - Equation (i) , We get

$\to$ ( a + 20d ) - ( a + 4d ) = 42 - 10

$\to$ 20d - 4d = 32

$\to$ 16d = 32

$\to$ d = 32 / 16

$\to$ d = 2

Put the value of d = 2 in eq (i)

$\to$ a + 4(2) = 10

$\to$ a + 8 = 10

$\to$ a = 10 - 8

$\to$ a = 2

We have to find 31st term So ,

$\to$ 31st term = a + ( 31 - 1 )d

$\to$ 31st term = a + 30d

$\to$ 31st term = 2 + 30(2)

$\to$ 31st term = 60 + 2

$\to$ 31st term = 62