Question 22

# Let $$a_{n}$$ and $$b_{n}$$ be two sequences such that $$a_{n}=13+6(n-1)$$ and $$b_{n}=15+7(n-1)$$ for all natural numbers n. Then, the largest three digit integer that is common to both these sequences, is

Solution

It is given that $$a_{n}=13+6(n-1)$$, which can be written as $$a_n=13+6n-6\ =\ 7+6n$$

Similarly, $$b_{n}=15+7(n-1)$$, which can be written as $$b_n=15+7n-7\ =\ 8+7n$$

The common differences are 6, and 7, respectively, The common difference of terms that exists in both series is l.c.m (6, 7) = 42

The first common term of the first two series is 43 (by inspection)

Hence, we need to find the mth term, which is less than 1000, and the largest three-digit integer, and exists in both series.

$$t_m=a+\left(m-1\right)d\ <\ 1000$$

=> $$43+\left(m-1\right)42\ <\ 1000$$

=> $$\left(m-1\right)42\ <\ 957$$

=> m-1 < 22.8 => m < 23.8 => m = 23

Hence, the 23rd term is $$43+22\times\ 42\ =\ 967$$

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