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There are two arithmetic progression, 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 in common in the first n terms of A1 and A2, if the sum of the nth terms of A1 and A2 is equal to 6,000?
Asked by rushabhjain.av | 08 Feb, 2019, 03:55: PM
nth term of Arithmetic progression A1 = 3 + (n-1)6
nth term of Arithmetic progression A2 = 5 + (n-1)8

Sum of both nth terms = 14n - 6 = 6000   or  n = 429

let pth term of A1 equals qth term of A2 :  then we have,   3+(p-1)6 = 5+(q-1)8   or  3p = 4q

from the above equation, we know that 4-th term of A1 or multiples of 4-th term of A1 is equal to
3-rd term of A2 or respective muliples of 3-rd term of A2

if last multiple of 4-th term in first 429 terms of A1 is 428,  then this number is 107-th multiple of 4-th term

Corresponding 107-th multiple of 3-rd term in A2 is 321

hence we have, 107 common terms of A1 and A2 in first 429 terms.

Answered by Thiyagarajan K | 10 Feb, 2019, 03:04: PM

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