find the number of terms comon to two A.P.'s : 3,7,11,.....,407 and 2,9,16,....,709.
Asked by gurpreet kaur | 16th Jun, 2013, 08:13: PM
For the first AP, a =3, d = 4
Hence, any nth term would be given by 3+(n-1)4 = 4n-1
Also, since 407 is the last term, so, 407 = 4n-1 i.e. maximum value of n can be 102
For the second AP, a = 2, d = 7
Hence any mth term would be given by 2+(m-1)7 = 7m - 5
Also, since 709 is the last terms, so, 709 = 7m-5 i.e. maximum value of m can be 102
To find the terms common to both the APs, we can equate the nth term of first AP to the mth term of the second AP
4n -1 = 7m-5
4n = 7m-4
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, ......100
So, the number of terms common to the 2 APs would be 25.
Answered by | 16th Jun, 2013, 08:44: PM
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