There are two arithmetic progressions,A1 and A2 whose first terms are 3 and 5 respectively and whose common difference are 6 and 8 respectively. How many terms of the series are common in the first n terms of A1 and A2 if the sum of the nth term of A1 and A2 is equal to 6000 ??/

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Solution

Dear student
$Given the arithmatic progression A_{1} \to 3, 9, 15, 21 . . . and A_{2} \to 5, 13, 21, 29 We can see that first term common between the two series is 21 . Since the common difference is 6 and 8 respectively So, 6000 = a_{1} + a_{2} + (n - 1) (d_{1} + d_{2}) 6000 = (3 + 5) + (n - 1) (6 + 8) 6000 = 8 + 14 n - 14 14 n - 6 = 6000 so n = \frac{6006}{14} = 429 A of A_{1} = (3 + (429 - 1) \times 6) = 2571 A of A_{2} = (5 + (429 - 1) \times 8) = 3427 So all the terms common to A_{1} and A_{2} will be less than 2571 . So the number of terms common to A_{1} and A_{2} is same as the number of terms in A_{3} less than or equal to 2571 ie [\frac{2571 - 21}{24}] + 1 - [\frac{2550}{24}] + 1 = 107$
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