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प्रश्न
Find the common difference of the Arithmetic progression
`1/a, (3 - a)/(3a), (3 - 2a)/(3a),"......"(a ≠ 0)`
योग
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उत्तर
Given: The AP: `1/a, (3 - a)/(3a), (3 - 2a)/(3a),"......"(a ≠ 0)`. Common difference d = a2 – a1 by definition of AP.
Step-wise calculation:
1. `T_1 = 1/a`.
2. `T_2 = (3 - a)/(3a)`
d = T2 − T1
= `(3 - a)/(3a) - 1/a`
= `(3 - a)/(3a) - 3/(3a)`
= `(3 - a - 3)/(3a)`
= `(-a)/(3a)`
= `-1/3`
3. Verify with T3:
`T_3 = (3 - 2a)/(3a)`
`T_3 − T_2 = ((3 - 2a) - (3 - a))/(3a)`
= `(-a)/(3a)`
= `-1/3`
The common difference is d = `-1/3`.
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