Question

If the 7th term of A.P is1/9 and 9th term is1/7, find its 63th term

Answer 1

A7 = 1/9

a + 6d = 1/9

A9 = 1/7

a + 8d = 1/7

solving two equations

a + 6d = 1/9

a + 8d = 1/7

___________

– 2d = 1/9 – 1/7

– 2d = – 2 / 63

d = 1 / 63

put d = 1/63 in 1

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

a = 1/9 – 6/63

a = ( 7 – 6 ) / 63

a = 1 /63

Hence a = d

A63 = a + 62d

= d + 62d

= 63d

= 63 × 1 /63

= 1

Hence 63rd term is 1

Answer 2

ANSWER

The 63rd term is 1

GIVEN

• The 7th term of an AP is 1/9
• and 9th term is 1/7

SOLUTION

Let us consider the first term of an AP be a and common difference be d

We are given ,

7th term = 1/9

$\sf \implies a + (7 - 1)d = \dfrac{1}{9} \\ \\ \implies \sf a + 6d = \dfrac{1}{9} \: \: .........(1)$

and

9th term = 1/7

$\sf \implies a + (9 - 1)d = \dfrac{1}{7} \\ \\ \implies \sf a + 8d = \dfrac{1}{7} \: \: ........(2)$

Subtracting (1) from (2) we have :

$\sf \implies a + 8d - (a + 6d) = \dfrac{1}{7} - \dfrac{1}{9} \\ \\ \implies \sf a + 8d - a - 6d = \dfrac{9 - 7}{63} \\ \\ \sf\implies2d = \dfrac{2}{63} \\ \\ \implies \sf d = \dfrac{1}{63}$

Thus , common difference is 1/63

Putting the value of d in (2)

$\sf \implies a + 6 \times \dfrac{1}{63} = \frac{1}{9} \\ \\ \implies \sf a = \dfrac{1}{9} - \dfrac{6}{63} \\ \\ \sf \implies a = \dfrac{7 - 6}{63} \\ \\ \implies \sf a = \dfrac{1}{63}$

Thus , the first term is 1/63

Now the 63th term will be :

$\sf \implies t_{63} = \dfrac{1}{63} + (63 - 1) \dfrac{1}{63} \\ \\ \implies\sf t_{63} = \dfrac{1}{63} + \frac{62}{63} \\ \\ \implies\sf t_{63} = \dfrac{1 + 62}{63} \\ \\ </p><p> \sf\implies t_{63} = \dfrac{63}{63} \\\\ \implies\sf t_{63} = 1$

Hence , the 63rd term is 1