Question-94960

Question Number 94960 by bshahid010@gmail.com last updated on 22/May/20

Answered by Kunal12588 last updated on 22/May/20

2 + (n - 1) (3) = 2 + (n_{1} - 1) (5)

\Rightarrow 3 n - 1 = 5 n_{1} - 3

\Rightarrow 5 n_{1} - 3 n = 2

n_{1} = 1 + 3 t

n = 1 + 5 t

(\begin{matrix} n \\ n_{1} \end{matrix}) = (\begin{matrix} 1 & 6 & 11 & 16 \\ 1 & 4 & 7 & 10 \end{matrix})

4 c o m m o n t e r m s

Answered by mr W last updated on 22/May/20

A_{n} = 2 + 3 n = 2 + 3 \times 5 m ⩽ 59

B_{k} = 2 + 5 k = 2 + 5 \times 3 m ⩽ 59

m ⩽ \frac{59 - 2}{3 \times 5} = 3.8

\Rightarrow 0 ⩽ m ⩽ 3

\Rightarrow 4 c o m m o n t e r m s :

2, 17, 32, 47

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