Question

Find the number of terms common to the two $A.P{.}^{\prime }s3,7,11,$______$407$ and $2,9,16$,______$709$.

Answer 1

It is easy to observe that both the series consist of $102$ terms. Let $T_p=3+4(p-1)=4p-1$ and $T_q=2+7(q-1)=7q-5$ 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., $4p-1=7q-5$ or $4(p+1)=7q$ $\left(1\right)$
Now p and q are $+$ive integers and hence from $\left(1\right)$ we conclude that q is multiple of $4$ and so let q$=4s$ and as q lies between $1$ and $102$, therefore s lies between $1$ and $25$.
$\therefore \frac{p+1}{7}=\frac{q}{4}=\lambda$
$p+1=7\lambda$ and $q=4\lambda$
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.