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Question Numbers: 136-138Two number series are given below, each following a different pattern:
Series A:
5, 3, 16, 97, P, 5441
Series B:
Q,(Q−400),(3.5P+172),2296,2100,1956,RBased on the given series, answer the following questions:
Solution:
Concept:Find the missing terms
P,
Q, and
R by identifying the pattern in each series, then compute
Q−R.
Explanation:For Series A, starting from the second term, each term is obtained by multiplying the previous term by consecutive numbers
5,6,7,8 and adding
1.
3×5+1=1616×6+1=9797×7+1=680680×8+1=5441Hence,
P=680.
Using
P, the third term of Series B is:
3.5×680+172=2380+172=2552So Series B becomes:
Q,
Q−400,
2552,
2296,
2100,
1956,
R.
Check the differences between known terms:
2552−2296=256=1622296−2100=196=1422100−1956=144=122The differences are squares of even numbers in descending order.
So the next difference before
R is
102:
R=1956−102=1956−100=1856For the first gap, the difference is
202, and for the second gap it is
182:
Q−(Q−400)=400=202Q−400=2552+182=2552+324=2876Q=2876+400=3276Therefore:
Q−R=3276−1856=1420Answer:1420The correct option is B.
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