Here is the question –
Certain numbers appear in both arithmetic progressions 17, 21, 25, …
and 16, 21, 26, … . Find the sum of first 100 numbers appearing in
both progressions.
The solution given is –
(1) Denoting the nth and mth terms of the 2 progressions by Tn
and Tm', we have(2) Tn = 17 + (n – 1) . 4 = 4n + 13 and
(3) Tm' = 16 + (m – 1) . 5 = 5m + 11
(4) For common terms, we must have
(5) Tn = Tm' => 4n + 13 = 5m + 11 => 5m = 2 . (2n + 1)
(6) This shows that 2n + 1 = 5k, k = 1, 3, 5, …
(7) Hence, the common terms are given by –
(8) T(2k)' = 5 . 2k + 11 = 10k + 11, k = 1, 3, 5, …
(9) Hence, sum of first 100 common terms = 21 + 41 + 61 + … =
(100/2)(2×21+(100-1).20) = 101100
I can understand clearly till line (5) in the solution, but after that, whatever is done confuses me. Could anyone explain how to go about such a problem?
Best Answer
The first sequence has the form $13+4 m$, while the second has the form $11+5 n$. These are equal when $5 n-4 m = 2$, or, for any nonnegative integer $k$, $m=2+5 k$ and $n=2+4 k$ for members of both sequences. Thus, the form of the sequence is $a_k = 11+5 (2+4 k) = 21+20 k$. The sum of the first 100 of these terms is
$$\sum_{k=0}^{99} (21+20 k) = 100 \cdot 21 + \frac12 (20) (100)(99) = 101100$$
ADDENDUM
Looks like you were confused about how to express the common term $a_k$. Here, I just made a table of ordered pairs $(m,n)$ that satisfied $5 n-4 m=2$, which isn't hard. Once you see the pattern, the above expressions jump out.