Question

if the 7th term of an A.P. is 1 by 9 and its ninth term is 1 by 7, find its 63rd term

Answer 1

Heya !!!

7th term = 1/9

A + 6D = 1/9

9(A + 6D) = 1

9A + 54D = 1 ---------(1)

and,

9th term = 1/7

A + 8D = 1/7

7(A + 8D) = 1

7A + 56D = 1 ----------(2)

From equation (1) we get,

9A + 54D = 1

9A = 1-54D

A = 1 -54D/9 -----------(3)

Putting the value of A in equation (2)

7A + 56D = 1

7 × (1-54D/9) + 56D = 1

7 - 378D/9 + 56D = 1

7 - 378D + 504D = 1 × 9

126D = 9 -7

126D = 2

D = 2/126 => 1/63

Putting the value of D in equation (3)

A = 1-54D/9 => 1 - 54 × 1/63 /9

A = 1 - 54/63 /9 = 63-54/63 /9. = 9 /63 /9

A = 9/63 × 1/9

A = 1/63

Therefore,

63 Rd term = A+ 62D = 1/63 + 62 × 1/63

=> 1/63 + 62/63


=> 63/63 = 1


Hence,


63Rd term of AP is 1.



HOPE IT WILL HELP YOU...... :-)

Answer 2

7th term = a + 6d

1/9 = a + 6d

a = 1/9 - 6d ----1equation

×××××××××××××××××××

9th term = a + 8d

1/7 = a + 8d

Putting the value of a from 1equation,

1/7 = 1/9 - 6d + 8d

1/7 - 1/9 = 2d

(9-7)/63 = 2d

2/63 = 2d

1/63 = d

Putting the value of d in 1equation,

a = 1/9 - 6(1/63)

=> a = 1/9 - 2/21

=> a = (7 - 6)/63

=> a = 1/63



======================

Then,

63th term = a + 62d

=> 1/63 + 62/63

=> (1+ 62)/63

=> 63/63

=> 1




I hope this will help you

(-: