Question

How many terms are identical in an AP's 2,5,8,......upto 50 terms and 3,5,7,9....upto 60terms?

Answer 1

AP.1 $\Rightarrow$ 2, 5, 8 . . . . . 50 terms
$T_{50}$ = 2 + 49 $\times$ 3
= 2 + 147
= 149
$\therefore$ 2, 5, 8, . . . . . 149
AP.2 $\Rightarrow$ 3, 5, 7, 9, . . . . . 60 terms
$T_{60}$ = 3 + (60 - 1)2
= 3 + 59 $\times$2
= 3 + 118
= 121
3, 5, 7. . . . . . . 121
Common difference in $A{P}_1$ = 3
Common difference in $A{P}_2$ = 2
Let $A{P}_3$ = A.P with common terms of $A{P}_1\&A{P}_2$
Here, d = LCM of 2 & 3 = 6
First term of $A{P}_3=5$ (1st common term)
$\therefore A{P}_3$ = 5, 11, 17, 23 . . . . . .
$T_n$ = 5 + (n - 1)6
Let term of $A{P}_2$ = 121
$\therefore A{P}_3$ should be less than 121.
So, we have to check for nearest term to 121.
When you see $T_n(=5+kd)$
Carefully, where k = (n - 1), you can say that $T_n)$ is such that, when divided by 6 leaves a reminder of 5.
$\therefore$ Probable values can be 125, 119
Since, our last term should be less than 121.
$\therefore$ Last term = 119
$\therefore$ 5 + (n - 1)6 = 119
$\Rightarrow$ (n - 1)6 = 114
n - 1 = 19
$\Rightarrow$ n = 20
$\therefore$ 20 terms will be identical
5, 11, 17, 23, 29,
35, 41, 47, 53, 59,
65, 71, 77, 83, 89,
95, 101, 107, 113, 119