## Monday, 7 November 2011

### Progressions-7 (CAT-2008)

The number of common terms in the two sequences 17,21,25,….,417 and 16,21,26,….466 is
(1)78          (2) 19              (3)20                    (4)77               (5) 22
Solution follows here:
Solution:
Both the given series’ are in A.P.
Number of terms in the 1st series: 417 = 17+(n1-1)4 => n1-1 = 400/4 = 100 => n1=101
Number of terms in the 2nd series: 466 = 16+(n2-1)5 => n2-1 = 450/5 = 90 => n2=91
To find number of common terms:
Let m’th term of the first sequence = n’th term of the second sequence
Using formula for n’th term in an AP with initial term ‘a’ and common difference ‘d’,
tn = a+(n-1)d,
17+(m-1)4 = 16+(n-1)5 => 4m-4+17 = 5n-5+16 => 4m+2 = 5n
=> we can observe that, m’th term of the first sequence is equal to the n’th term of the second sequence, for
m=2, n=2;
m=7, n=6;
m=12, n=10;
… and so on…
m=97, n=78;
2nd, 7th, 12th, …..97th terms of the first series are respectively equal to the 2nd, 6th, 10th, …..78th terms of the second series. These are the common terms.
Number of common terms: 97 = 2+(x-1)5 => x-1 = 95/5 = 19=> x=20