Question

Find the number of terms common to the two A.P' s $3,7,\mathrm{11....}407$ and $2,9,16,....709$.

Answer 1

It is easy to observe that both the series consist of $102$ terms. Let
$T_p=3+4(p-1)=4p-1and{T}_q=2+7(q-1)=7q-5$
be the general terms of the two series where both p and q lie between $1$ and $102$. We have to find the values of p and q for which $T_p=T_q$
i.e., $4p-1=7q-5or4(p+1)=7q...\left(1\right)$
Now p and q are positive integers and hence from (1) we conclude that q is multiple of $4$ and so let $q=4s$ and as q lies between $1$ and $102$, therefore it lies between $1$ and $25$.
$\therefore \frac{p+1}{7}=\frac{q}{4}=\lambda$
$p+1=7\lambda andq=4\lambda$
both p and q vary from $1$ to $102$
$\therefore \lambda$ varies from $1$ to $14$ or from $1$ to $25$
Hence we choose $\lambda$ to vary from $1$ to $14$. Thus there are only $14$ common terms.
$T_p=4p-1=4(7\lambda -1)-1=28\lambda -5$
Put $\lambda =1,2,3,\cdots ,14$ and common terms are $23,51,79,...$