Question

If the seventh term of an A.P. is $\frac{1}{9}$ and its ninth term is $\frac{1}{7}$, find its (63)rd term.

Answer 1

Answer:

Its (63)rd term is 1.

Step-by-step explanation:

Given :  

a7 = 1/9 and a9 = 1/7  

By using the formula , nth term ,an = a + (n -1)d

Case 1 :  

a7 = a + (7 - 1) d  

1/9 = a + 6d

a + 6d  = 1/9 ……….(1)

Case 2 :  

a9 = a + (9 - 1) d  

1/7 =  a + 8d

a + 8d  =  1/7 ………(2)

On subtracting equation (2) form (1),

a + 8d = 1/7  

a + 6d = 1/9

(-)  (-)   (-)  

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2d = 1/7 - 1/9

2d = (9 - 7)/63

2d = 2/63

d = 2/63 × ½

d = 1/63

On putting the value of d = 1/63 in equation (1),

a + 6d = 1/9

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

a + 6/63  = 1/9

a = 1/9 - 6/63

a = (7 - 6)/63

a = 1/63

For 63rd term :  

a63 = a + (63 - 1) d

a63 = a + 62d

a63 = 1/63 + 62 × 1/63

a63 = 1/63 + 62/63

a63 = (1 + 62)/63

a63 = 63/63  

a63 = 1  

Hence, its (63)rd term is 1.

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Answer 2

Step-by-step explanation:

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