Find the number of terms common to the two A,P's 3,7,11,.....,407 and 2,9,16,..........,709
A
12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
14
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
25
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
28
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C25 The given terms of AP are 3,7,11.........407 and 2,916,......70
Now,
Let mth term at first sequence be same as the nth terms at the second sequence Then tm and tn
∵3+(m−1)×4=2+(n−1)7
⇒4m−1=n−5
⇒4m−7n=−5+1
⇒4m=7n−4
⇒m=7n−44
Now, since n is a whole number, so 7m-4 needs to be divisible by 4. So, then, m can be equal to all multiples of 4 till 102 i.e. 4, 8, 12, 16, 20, 24, ......100So, the number of terms common to the 2 APs would be 25.