Question

If the seventh term of an A.P. is 1/9 and its ninth term is 1/7, find its (63)rd term.

Answer 1

Answer:

t₆₃=1

Step-by-step explanation:

• tₙ=a+(n-1)d
• Given: t₇=1/9 and t₉=1/7
• t₇=1/9=a+6d   ...(1)  
• t₉=1/7=a+8d   ...(2)
• solving equations (1) and (2) simultaneously we get,
• d=1/63
• substituting value of d in equation (1), a=1/63=d
• Therefore the 63rd term is t₆₃=a+62d
• t₆₃=63d      ....(a=d)
• t₆₃=63/63=1    ....(d=1/63)
• t₆₃=1

Answer 2

Answer:

The value of 63rd terms is 1  

Step-by-step explanation:

Given as :

For Arithmetic progression

The seventh term of an A.P = $\dfrac{1}{9}$

The ninth term of an A.P = $\dfrac{1}{7}$

Since nth term of an A.P is written as

$t_n$ = a + ( n - 1 ) d

where a is first term

And d is common difference

So, For n = 7

$t_7$ = a + ( 7 - 1 ) d

i.e  $\dfrac{1}{9}$  =  a + 6 d

Or, 9 a + 54 d = 1                    .......A

Again

For n = 9

$t_9$ = a + ( 9 - 1 ) d

i.e  $\dfrac{1}{7}$  =  a + 8 d

Or, 7 a + 56 d = 1                     .......B

Solving eq A and B

7 ( 9 a + 54 d ) - 9 ( 7 a + 56 d ) = 7 × 1 - 9 × 1

Or, (63 a - 63 a) + ( 378 d - 504 d) = - 2

Or,  0 + 126 d = 2

∴            d = $\dfrac{2}{126}$

i.e          d = $\dfrac{1}{63}$

So, common difference = d = $\dfrac{1}{63}$

Put the value of d in eq A

9 a + 54 × $\dfrac{1}{63}$ = 1  

Or, 9 a = 1 - $\dfrac{54}{63}$

Or, 9 a = $\dfrac{9}{63}$

∴    a = $\dfrac{1}{63}$

So, first term = a = $\dfrac{1}{63}$

Now, For 63rd term

i.e  $t_6_3$ = $\dfrac{1}{63}$ + ( 63 - 1 ) × $\dfrac{1}{63}$

Or, $t_6_3$ = $\dfrac{1}{63}$ + $\dfrac{62}{63}$

∴     $t_6_3$  = $\dfrac{1 + 62}{63}$  = $\dfrac{63}{63}$

ie.    $t_6_3$ = 1

So, The value of 63rd terms = $t_6_3$ = 1

Hence, The value of 63rd terms is 1   . Answer