Question

The sum of 4th and 8th term of an AP is 24 and the sum of the 6 and 10th term is 44. Find the nth term of an AP?

Answer 1

EXPLANATION.

The sum of 4th and 8th term of an Ap = 24

The sum of 6th and 10th term of an Ap = 44

To find the Nth term of an Ap.

Formula of Nth term of an Ap.

=> An = a + ( n - 1 ) d

=> 4th and 8th term = 24

=> a + 3d + a + 7d = 24

=> 2a + 10d = 24

=> a + 5d = 12 .....(1)

=> 6th and 10th term = 44

=> a + 5d + a + 9d = 44

=> 2a + 14d = 44

=> a + 7d = 22 ......(2)

From equation (1) and (2)

we get,

=> -2d = -10

=> d = 5

put the value of d = 5 in equation (1)

we get,

=> a + 5(5) = 12

=> a + 25 = 12

=> a = -13

Therefore,

Nth term of an Ap

=> An = -13 + ( n - 1 ) 5

=> An = -13 + 5n - 5

=> An = 5n - 18

Nth term of an Ap = 5n - 18.

Answer 2

Step-by-step explanation:

$\bf{\underline{\underline\red{Given:-}}}$

• The sum of 4th and 8th term of an AP is 24.

• The sum of 6th and 10th term is 44.

$\bf{\underline{\underline\blue{To\:Find:-}}}$

• The nth term of the AP.

$\bf{\underline{\underline\green{Solution:-}}}$

As we know that:-

The nth term of the AP is given by the formula.

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

Here:-

• a = first term

• n = number of terms

• d = common difference.

The 4th term:-

$\rm \implies a_{4} = a + (4 - 1)d$

$\rm \implies a_{4} = a + 3d$

The 8th term:-

$\rm \implies a_{8} = a + (8- 1)d$

$\rm \implies a_{8} = a + 7d$

The sum of 4th and 8th term of an AP is 24.

$\rm \implies a_{4} + a_{6} = 24$

$\rm \implies (a + 3d) + (a + 7d)= 24$

$\rm \implies a + 3d + a + 7d= 24$

$\rm \implies 2a + 10d= 24$

Dividing the equation by 2

$\rm \implies a + 5d= 12......(i)$

The 6th term:-

$\rm \implies a_{6} = a + (6- 1)d$

$\rm \implies a_{6} = a + 5d$

The 10th term:-

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

$\rm \implies a_{10} = a + 9d$

The sum of 6th and 10th term of an AP is 44.

$\rm \implies a_{6} + a_{10} = 44$

$\rm \implies (a + 5d) + (a + 9d)= 44$

$\rm \implies a + 5d + a + 9d= 44$

$\rm \implies 2a + 14d= 44$

Dividing the equation by 2

$\rm \implies a + 7d= 22......(ii)$

Subtracting equation (i) from (ii)

$\rm \implies a + 7d - (a + 5d)= 22 - 12$

$\rm \implies a + 7d - a - 5d= 10$

$\rm \implies 2d= 10$

$\rm \implies d= 5$

Substituting d = 5 in equation (i)

$\rm \implies a + 5d= 12$

$\rm \implies a + 5(5)= 12$

$\rm \implies a + 25= 12$

$\rm \implies a = 12 - 25$

$\rm \implies a = - 13$

We have:- a = -13 and d = 5

The nth term:-

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

$\rm \implies a_{n} = - 1 3+ (n - 1)5$

$\rm \implies a_{n} = - 1 3+ 5n -5$

$\rm \implies a_{n} = 5n - 18$

$\underline{\boxed{ \purple{\rm \therefore The \: nth \: term \: of \: the \: AP = 5n - 18}}}$